Free math worksheets

- Both arc-intersection points have the same distance to \(A\) and \(B\). - A line through two points on the perpendicular-bisector locus is the desired line.

1. Each intersection point is equidistant from \(A\) and \(B\). 2. Two distinct points equidistant from \(A\) and \(B\) determine the perpendicular bisector of \(\overline{AB}\). 3. Therefore, draw line \(CD\).

Draw line \(CD\).

- Translate “the same distance from \(A\) and \(B\)” into a locus statement. - Recall which line consists exactly of points equidistant from a segment's endpoints.

1. The perpendicular-bisector theorem says that every point equidistant from \(A\) and \(B\) lies on the perpendicular bisector of \(\overline{AB}\). 2. Since \(PA=PB\), point \(P\) lies on that perpendicular bisector.

The perpendicular bisector of \(\overline{AB}\).

- The point \(F\) was created to lie symmetrically with respect to the two sides of the angle. - The final construction line must pass through the original vertex.

1. The equal-radius arcs make \(DF=EF\). 2. Because \(BD=BE\) from the first arc and \(BF\) is shared, the two construction triangles are congruent. 3. Therefore, the ray through \(B\) and \(F\) is the angle bisector.

Draw ray \(\overrightarrow{BF}\).

- Identify what equal distances from \(A\) and \(B\) tell you about both \(P\) and \(Q\). - The segment \(\overline{AB}\) lies on \(\ell\).

1. Since \(PA=PB\), point \(P\) lies on the perpendicular bisector of \(\overline{AB}\). 2. Since \(QA=QB\), point \(Q\) also lies on the perpendicular bisector of \(\overline{AB}\). 3. Therefore, line \(PQ\) is that perpendicular bisector and is perpendicular to \(\ell\) through \(P\).

It constructs the line perpendicular to \(\ell\) through \(P\).

- How large must the compass radius be for the arcs to intersect twice? - Where does a perpendicular bisector cross the original segment? - Compare the distances from any point on the perpendicular bisector to the endpoints.

1. Draw \(\overline{AB}=7.6\,\text{cm}\). 2. Open the compass to a radius greater than \(3.8\,\text{cm}\). With center \(A\), draw arcs above and below the segment. Without changing the compass width, repeat with center \(B\). 3. Draw the line through the two arc intersection points. This line is the perpendicular bisector of \(\overline{AB}\). 4. The midpoint divides the segment into two parts of length \(7.6\div2=3.8\,\text{cm}\). 5. Every point on the perpendicular bisector is equidistant from \(A\) and \(B\).

a) Draw equal-radius arcs from \(A\) and \(B\) with radius greater than \(3.8\,\text{cm}\), then draw the line through their two intersections. b) Each part is \(3.8\,\text{cm}\). c) Every point on the perpendicular bisector is the same distance from \(A\) as from \(B\).

- What happens to an angle measure when the angle is bisected? - Repeat the same construction on one of the smaller angles. - Track the angle created after each construction.

1. To construct \(w_1\), draw an arc centered at the vertex that intersects both sides of the angle. From those two intersection points, draw equal-radius arcs that intersect inside the angle. Draw a ray from the vertex through that intersection. 2. The first bisection creates two angles of measure \(74^\circ\div2=37^\circ\). 3. Apply the same construction to one \(37^\circ\) angle to create \(w_2\). 4. The smallest angle measures \(37^\circ\div2=18.5^\circ\).

a) Construct \(w_1\) using equal-radius arcs from points on the two sides of the angle. b) Repeat the angle-bisector construction on one \(37^\circ\) angle to construct \(w_2\). c) The smallest angle measures \(18.5^\circ\).

- A compass-and-straightedge angle bisector is located from two equal arcs drawn from points on the angle's sides. - After each step, identify the smaller angle adjacent to \(\overrightarrow{OA}\). - The numerical measures provide a check on the construction sequence, not a substitute for describing it.

1. Draw an arc centered at \(O\) that meets both sides of \(\angle AOB\). From those two intersection points, draw equal-radius arcs that cross inside the angle. Draw the ray from \(O\) through their intersection; this is the first angle bisector. 2. To continue, bisect the smaller angle between \(\overrightarrow{OA}\) and the newest bisector ray, using the same type of arc-intersection construction each time. 3. The successive angle measures are \(45^\circ\), \(22.5^\circ\), \(11.25^\circ\), and \(5.625^\circ\). 4. Therefore, four total bisections are required before the angle is smaller than \(10^\circ\).

a) Draw an arc centered at \(O\) that meets both sides, then draw equal-radius arcs from those two points and draw the ray through their intersection. b) Repeatedly bisect the smaller angle between \(\overrightarrow{OA}\) and the newest bisector ray; four total bisections are needed. c) The first two measures are \(45^\circ\) and \(22.5^\circ\), and the first resulting angle below \(10^\circ\) is \(5.625^\circ\).

- Identify the two triangles on opposite sides of the proposed bisector. - Translate each equal-radius arc into an equal-length statement. - Determine what the two triangles share.

1. Since \(A\) and \(B\) lie on the same arc centered at \(V\), \(VA=VB\). 2. Since \(C\) lies on arcs of equal radius centered at \(A\) and \(B\), \(AC=BC\). 3. Segment \(\overline{VC}\) is common to \(\triangle AVC\) and \(\triangle BVC\). 4. Therefore, \(\triangle AVC\cong\triangle BVC\) by SSS. 5. Corresponding angles \(\angle AVC\) and \(\angle CVB\) are congruent, so \(\overrightarrow{VC}\) bisects \(\angle AVB\).

The equal-radius arcs give \(VA=VB\) and \(AC=BC\), while \(VC\) is shared. Thus \(\triangle AVC\cong\triangle BVC\) by SSS, so \(\angle AVC\cong\angle CVB\). Therefore, \(\overrightarrow{VC}\) is the angle bisector.

- Use the fact that \(A\) and \(B\) lie on one circle with center \(P\). - Connect equal distances from two endpoints to a familiar locus. - Relate the segment being bisected to the original line.

1. Points \(A\) and \(B\) lie on the same circle centered at \(P\), so \(PA=PB\). 2. A point equidistant from the endpoints of a segment lies on that segment's perpendicular bisector. Therefore, \(P\) lies on the perpendicular bisector of \(\overline{AB}\). 3. Segment \(\overline{AB}\) lies on line \(\ell\). 4. The perpendicular bisector of \(\overline{AB}\) is perpendicular to \(\overline{AB}\), so it is perpendicular to \(\ell\). 5. Thus it is the line through \(P\) perpendicular to \(\ell\).

Since \(PA=PB\), point \(P\) lies on the perpendicular bisector of \(\overline{AB}\). Because \(\overline{AB}\) lies on \(\ell\), that bisector is perpendicular to \(\ell\). Therefore, it is the required perpendicular through \(P\).

- Treat the two adjacent angles as two separate angle-bisector constructions. - The original angles form a linear pair. - Express the angle between the constructed bisectors in terms of half of each original angle.

1. For \(\angle AVB\), draw an arc centered at \(V\) meeting both sides; from the two intersection points draw equal-radius arcs that meet inside the angle, then draw the ray from \(V\) through that intersection. 2. Repeat the same construction independently for \(\angle BVC\). 3. Let \(m\angle AVB=\alpha\). Because the angles form a linear pair, \(m\angle BVC=180^\circ-\alpha\). 4. The angle between the two constructed bisectors is \(\frac{\alpha}{2}+\frac{180^\circ-\alpha}{2}=90^\circ\). 5. Therefore, the two constructed bisectors are perpendicular.

Construct each internal bisector by the standard equal-arc intersection procedure. Since the original angles are supplementary, the angle between their bisectors is \(\frac{\alpha}{2}+\frac{180^\circ-\alpha}{2}=90^\circ\); hence the bisectors are perpendicular.

- Use the given segment itself as one compass opening from both endpoints. - Think about what is true of every intersection of two equal-radius circles centered at \(A\) and \(B\). - Two distinct points on the same locus determine its line.

1. Set the compass opening to \(AB\). Draw a circle centered at \(A\) and a circle centered at \(B\) with that same radius. 2. The two circles meet at two points on opposite sides of \(\overline{AB}\); name them \(C\) and \(D\). Draw line \(CD\). 3. Because \(C\) lies on both circles, \(CA=CB\). Because \(D\) lies on both circles, \(DA=DB\). 4. Thus both \(C\) and \(D\) lie on the perpendicular-bisector locus of \(\overline{AB}\). 5. The unique line through those two points is the perpendicular bisector of \(\overline{AB}\), so \(CD\) is that line.

Draw equal-radius circles centered at \(A\) and \(B\) with radius \(AB\); use their two intersections as \(C\) and \(D\), then draw \(CD\). Each intersection is equidistant from \(A\) and \(B\), so both lie on the perpendicular bisector. Therefore, \(CD\) is the perpendicular bisector of \(\overline{AB}\).

- Compare the two triangles formed by the center, the midpoint, and the chord endpoints. - Identify equal lengths coming from the circle and from the midpoint. - Use the relationship between the two adjacent angles at \(M\).

1. Radii \(\overline{OA}\) and \(\overline{OB}\) are congruent, so \(OA=OB\). 2. Since \(M\) is the midpoint of \(\overline{AB}\), \(AM=MB\). 3. Segment \(\overline{OM}\) is common to \(\triangle OMA\) and \(\triangle OMB\). 4. Therefore, \(\triangle OMA\cong\triangle OMB\) by SSS. 5. Corresponding angles \(\angle OMA\) and \(\angle OMB\) are congruent and form a linear pair. 6. Congruent supplementary angles each measure \(90^\circ\), so \(OM\perp AB\).

The triangles on either side of \(\overline{OM}\) are congruent by SSS. Their angles at \(M\) are congruent and supplementary, so each is \(90^\circ\). Therefore, \(OM\perp AB\).

- First identify the complete locus of points equidistant from \(A\) and \(B\). - Recall how to construct that locus from the two endpoints without measuring a midpoint. - The number of road solutions is the number of intersections of two lines.

1. Construct the perpendicular bisector of \(\overline{AB}\): draw equal-radius arcs centered at \(A\) and \(B\) with radius greater than half of \(AB\), then draw the line through the two arc intersections. 2. Mark its intersection with road line \(r\) as \(P\). 3. Every point on the perpendicular bisector is equidistant from \(A\) and \(B\), so \(PA=PB\). 4. Because \(r\) is neither parallel to nor coincident with the perpendicular bisector, the two lines meet at exactly one point, so \(P\) is unique. 5. If the lines are parallel, there is no such road point. If they coincide, every point on the road is equidistant from the towers.

Construct the perpendicular bisector of \(\overline{AB}\) with equal-radius arcs from \(A\) and \(B\), then take its intersection with \(r\) as \(P\). The stated line conditions make that intersection unique. A parallel road gives no solution; a coincident road gives infinitely many.

- Equal perpendicular distances to two intersecting lines characterize their angle-bisector locus. - The region containing \(X\) determines which ray of the locus is relevant. - For the construction, begin by marking the two sides with one arc centered at the vertex.

1. Points equidistant from two intersecting lines lie on one of their two angle-bisector lines. 2. Because \(X\) lies inside the specified obtuse region, ray \(\overrightarrow{VX}\) is the bisector ray for that obtuse angle rather than for either acute angle. 3. To construct the bisector from the angle alone, draw an arc centered at \(V\) that meets the two sides of the obtuse angle. 4. From those two intersection points, draw equal-radius arcs that meet inside the obtuse angle. Draw the ray from \(V\) through their intersection. 5. The standard construction produces the same angle-bisector ray identified by the equal-distance locus.

a) Equal distances place \(X\) on an angle-bisector line, and its location inside the obtuse region selects the obtuse-angle bisector, so \(\overrightarrow{VX}\) is that ray. b) Use an arc centered at \(V\) to mark the two sides, draw equal-radius arcs from those marks to meet inside the obtuse angle, and draw the ray from \(V\) through their intersection.

- Use the radius relationships in each circle separately. - Identify the locus of points equidistant from the chord endpoints. - Use the fact that the two centers are distinct.

1. Points \(A\) and \(B\) lie on the circle centered at \(O_1\), so \(O_1A=O_1B\). 2. Therefore, \(O_1\) lies on the perpendicular bisector of \(\overline{AB}\). 3. Points \(A\) and \(B\) also lie on the circle centered at \(O_2\), so \(O_2A=O_2B\). 4. Therefore, \(O_2\) lies on the same perpendicular bisector of \(\overline{AB}\). 5. The line through the two distinct points \(O_1\) and \(O_2\) is that perpendicular bisector.

Each center is equidistant from \(A\) and \(B\), so both centers lie on the perpendicular bisector of \(\overline{AB}\). Hence line \(O_1O_2\) is the perpendicular bisector of the common chord.

- Determine the relationship between two lines perpendicular to the same line. - Use the original pair of parallel lines for the other pair of opposite sides. - Identify the resulting quadrilateral and use one of its side properties.

1. Since \(\overline{AB}\perp\ell\) and \(\overline{CD}\perp\ell\), \(\overline{AB}\parallel\overline{CD}\). 2. Segments \(\overline{AC}\) and \(\overline{BD}\) lie on the parallel lines \(\ell\) and \(m\), so \(\overline{AC}\parallel\overline{BD}\). 3. Quadrilateral \(ACDB\) has two pairs of parallel opposite sides and a right angle, so it is a rectangle. 4. Opposite sides of a rectangle are congruent. Therefore, \(AB=CD\).

The app-created perpendiculars are opposite sides of rectangle \(ACDB\). Therefore, \(AB=CD\).

- Work with each chord independently before intersecting the results. - The center has an equal-distance relationship with both endpoints of any chord. - Nonparallel chords produce perpendicular bisectors with different directions.

1. Construct the perpendicular bisector of \(\overline{AB}\) using equal-radius arcs from \(A\) and \(B\). 2. Construct the perpendicular bisector of \(\overline{CD}\) the same way. 3. The center of a circle is equidistant from the endpoints of every chord, so it lies on both perpendicular bisectors. 4. Since the chords are nonparallel, their perpendicular bisectors are distinct and nonparallel, so they intersect at exactly one point. 5. The circle's center must lie at that unique intersection.

Construct the perpendicular bisectors of \(\overline{AB}\) and \(\overline{CD}\) with equal-radius arcs. Their unique intersection is the circle's center because the center is equidistant from the endpoints of each chord.

- Draw the radii to the two tangent points and identify the resulting right triangles. - Compare the two right triangles that share \(\overline{PO}\). - A congruence statement about the angles at \(P\) will locate the center relative to \(\angle TPU\).

1. Radii to points of tangency are perpendicular to the tangents, so \(OT\perp PT\) and \(OU\perp PU\). 2. Right triangles \(PTO\) and \(PUO\) share hypotenuse \(\overline{PO}\), and \(OT=OU\) because both are radii. 3. The triangles are congruent by HL. 4. Therefore, \(\angle TPO\cong\angle OPU\). 5. Hence ray \(\overrightarrow{PO}\) bisects \(\angle TPU\), so the center lies on its angle bisector.

The right triangles \(PTO\) and \(PUO\) are congruent by HL, so \(\angle TPO=\angle OPU\). Therefore, \(O\) lies on the angle bisector of \(\angle TPU\).

- Compare the two triangles formed by the center and the congruent chords. - Identify the equal radii and the shared segment. - Use the resulting triangle congruence to compare the two angles at \(A\).

1. In triangles \(AOB\) and \(AOC\), \(AB=AC\) by the given chord congruence. 2. Also, \(OB=OC\) because they are radii, and \(AO\) is shared. 3. Therefore, the triangles are congruent by SSS. 4. Corresponding angles \(\angle BAO\) and \(\angle OAC\) are congruent. 5. Hence ray \(\overrightarrow{AO}\) bisects \(\angle BAC\).

Triangles \(AOB\) and \(AOC\) are congruent by SSS, so \(\angle BAO=\angle OAC\). Therefore, \(\overrightarrow{AO}\) is the angle bisector.

- Relate the two pieces of \(\overline{BC}\) to the two sides adjacent to \(\angle A\). - The two pieces must have a fixed ratio and also add to the whole side. - Check that your two lengths add to \(12\,\text{cm}\).

1. By the Angle Bisector Theorem, \(\frac{BD}{DC}=\frac{AB}{AC}=\frac{10}{6}=\frac{5}{3}\). 2. Let \(BD=5k\) and \(DC=3k\). 3. Since \(BD+DC=BC=12\,\text{cm}\), \(8k=12\), so \(k=1.5\). 4. Therefore, \(BD=7.5\,\text{cm}\) and \(DC=4.5\,\text{cm}\).

\(BD=7.5\,\text{cm}\) and \(DC=4.5\,\text{cm}\).

- What does the altitude from the vertex of an isosceles triangle do to the base? - Use half of the base and the altitude to identify a right triangle. - The requested side is the hypotenuse of that right triangle.

1. In an isosceles triangle, the altitude from the vertex to the base also bisects the base. 2. Therefore, half of the base is \(5\,\text{cm}\). 3. One congruent side is the hypotenuse of a right triangle with legs \(5\,\text{cm}\) and \(6\,\text{cm}\). 4. By the Pythagorean theorem, the side length is \(\sqrt{5^2+6^2}=\sqrt{61}\,\text{cm}\).

\(\sqrt{61}\,\text{cm}\)

- An altitude is perpendicular to the line containing the opposite side, not necessarily to the side segment itself. - Use the obtuse angle at \(A\) to decide where the perpendicular feet fall. - Once two altitudes are built, recall the concurrency theorem for triangle altitudes.

1. For the altitude from \(B\), use the line containing \(AC\); because \(\angle A\) is obtuse, the perpendicular foot lies on the extension beyond \(A\). 2. For the altitude from \(C\), use the line containing \(AB\); its perpendicular foot also lies on the extension beyond \(A\). 3. Construct through \(B\) the line perpendicular to line \(AC\), and through \(C\) the line perpendicular to line \(AB\). 4. These two altitudes intersect outside the triangle at \(H\) because both relevant feet lie outside the opposite side segments. 5. The three altitudes of a triangle are concurrent, so the altitude through \(A\) also passes through \(H\). Therefore, two altitudes determine the orthocenter.

a) Extend the lines containing \(AC\) and \(AB\) beyond \(A\). b) Draw the perpendicular through \(B\) to line \(AC\) and the perpendicular through \(C\) to line \(AB\). c) They meet outside the triangle because their feet lie on side extensions. Altitude concurrency guarantees that their intersection \(H\) is the orthocenter.

- First reproduce the direction of \(\ell\) through \(P\). - Use perpendicular lines to create right angles at \(P\). - Count full lines rather than individual opposite rays.

1. The app creates line \(n\) through \(P\) perpendicular to \(\ell\). 2. It then creates line \(p\) through \(P\) perpendicular to \(n\). Therefore, \(p\parallel\ell\). 3. Lines \(p\) and \(n\) form four right angles at \(P\). The app bisects two adjacent right angles. 4. Each resulting bisector makes a \(45^\circ\) angle with \(p\), and therefore with \(\ell\), because \(p\parallel\ell\). 5. Bisectors of opposite right angles form the same full line. Thus the four bisector rays combine into exactly two distinct lines through \(P\).

The app creates a line through \(P\) parallel to \(\ell\) using two perpendiculars, then bisects two adjacent right angles at \(P\). The two resulting full lines are exactly the lines making an acute \(45^\circ\) angle with \(\ell\).

- Use the equal radii to compare each intersection point's distances from \(A\) and \(B\). - Recall the locus of points equidistant from the endpoints of a segment. - Use the midpoint property to divide the given length into two equal parts.

1. Because the two circles have the same radius, \(PA=PB\) and \(QA=QB\). 2. Therefore, both \(P\) and \(Q\) lie on the perpendicular-bisector locus of \(\overline{AB}\). The line through these two points, \(PQ\), is the perpendicular bisector of \(\overline{AB}\). 3. Since \(M\) is the midpoint of a \(9\,\text{cm}\) segment, \(AM=MB=\frac{9}{2}=4.5\,\text{cm}\). 4. The defining properties are that \(PQ\perp AB\) and that \(PQ\) passes through the midpoint \(M\) of \(\overline{AB}\).

a) Since \(PA=PB\) and \(QA=QB\), both \(P\) and \(Q\) lie on the perpendicular bisector of \(\overline{AB}\). Therefore, line \(PQ\) is that perpendicular bisector. b) \(AM=MB=4.5\,\text{cm}\). c) Line \(PQ\) is perpendicular to \(\overline{AB}\) and passes through its midpoint \(M\).

- Compare the distance between the centers with the sum of the two radii. - Think about when two congruent circles cross twice instead of touching once. - What is true about the distances from either intersection to the two centers?

1. Since \(UV=14\,\text{cm}\) and each radius is \(7\,\text{cm}\), the sum of the radii equals the distance between the centers. The circles are tangent and have only one intersection. 2. The common radius must be greater than \(7\,\text{cm}\) so the circles overlap and intersect at two points. 3. Each intersection point is the same distance from \(U\) and \(V\) because it lies on both circles with the same radius. 4. Every point equidistant from \(U\) and \(V\) lies on the perpendicular bisector of \(\overline{UV}\). Therefore, the line through the two intersections is that perpendicular bisector.

a) The circles are tangent because \(7+7=14\), so they provide only one intersection point. b) Use any common radius greater than \(7\,\text{cm}\). c) Both intersections are equidistant from \(U\) and \(V\), so the line through them is the perpendicular bisector of \(\overline{UV}\).

- Compare the two triangles formed on either side of \(\overline{PC}\). - Translate the point-placement and arc conditions into equal lengths. - Use both the equality and the sum of the two angles at \(P\).

1. The placement of \(A\) and \(B\) gives \(PA=PB\). 2. The equal-radius arcs give \(CA=CB\). 3. Segment \(\overline{PC}\) is common to \(\triangle APC\) and \(\triangle BPC\). 4. Therefore, \(\triangle APC\cong\triangle BPC\) by SSS. 5. Corresponding angles \(\angle APC\) and \(\angle CPB\) are congruent. 6. These adjacent angles form a linear pair, so their measures sum to \(180^\circ\). Congruent supplementary angles each measure \(90^\circ\). 7. Therefore, \(\overline{PC}\perp\ell\).

The equal lengths give \(\triangle APC\cong\triangle BPC\) by SSS. Thus \(\angle APC\cong\angle CPB\). Because they form a linear pair, each is \(90^\circ\), so \(PC\perp\ell\).

- Construct the minor-angle bisector before reasoning about the reflex angle. - The opposite ray of a constructed bisector lies on the same line through the vertex. - Compare each reflex-angle part with one half of the minor angle.

1. Draw an arc centered at \(V\) meeting the two sides of the minor angle. From those two intersection points, draw equal-radius arcs that meet inside the minor angle. Draw \(\overrightarrow{VW}\) through their intersection. 2. Extend line \(VW\) through \(V\) to form the opposite ray \(\overrightarrow{VW'}\). 3. Since \(\overrightarrow{VW}\) bisects the minor angle, each half has measure \(\frac{\theta}{2}\). 4. Each part of the reflex angle cut by \(\overrightarrow{VW'}\) has measure \(180^\circ-\frac{\theta}{2}\). 5. Those two measures are equal, so \(\overrightarrow{VW'}\) bisects the reflex angle.

Construct \(\overrightarrow{VW}\) by the standard equal-arc intersection method and extend it through \(V\). Each part of the reflex angle adjacent to \(\overrightarrow{VW'}\) measures \(180^\circ-\frac{\theta}{2}\), so the opposite ray is the reflex-angle bisector.

- Create two points on \(m\) that are automatically the same distance from \(A\). - A point equidistant from the endpoints of a segment lies on its perpendicular bisector. - After constructing the perpendicular foot, compare it with any oblique segment using a right triangle.

1. Draw a circle centered at \(A\) large enough to intersect line \(m\) at two points \(C\) and \(D\). Then \(AC=AD\). 2. Construct the perpendicular bisector of \(\overline{CD}\) using equal-radius arcs from \(C\) and \(D\). Because \(A\) is equidistant from \(C\) and \(D\), this perpendicular bisector passes through \(A\). 3. Let its intersection with \(m\) be \(B\). Since \(CD\) lies on \(m\), the constructed line \(AB\) is perpendicular to \(m\). 4. For any other point \(E\ne B\) on \(m\), triangle \(ABE\) is right at \(B\), so \(AE\) is the hypotenuse. 5. A hypotenuse is longer than either leg, so \(AE>AB\). Therefore, \(AB\) is the unique shortest segment from \(A\) to \(m\).

a) Intersect \(m\) with a circle centered at \(A\) to obtain \(C\) and \(D\), then construct the perpendicular bisector of \(\overline{CD}\). It passes through \(A\) and meets \(m\) at the required foot \(B\). b) For any other \(E\in m\), triangle \(ABE\) is right at \(B\), so \(AE\) is the hypotenuse and \(AE>AB\).

- A valid center must be exactly one radius from both given points. - Turn each distance condition into a circle centered at the corresponding endpoint. - After constructing the intersections, use the midpoint of \(\overline{AB}\) to form a right triangle.

1. Set the compass to \(5\,\text{cm}\). Draw a circle centered at \(A\) and a circle centered at \(B\) with that same radius. 2. Their two intersection points \(O_1\) and \(O_2\) are exactly the possible centers, because each is \(5\,\text{cm}\) from both \(A\) and \(B\). 3. The circles have two intersections because the center distance satisfies \(0<8<5+5\). 4. Their common chord is perpendicular to \(\overline{AB}\), so both centers lie on the perpendicular bisector. If \(M\) is the midpoint, then \(AM=4\,\text{cm}\). 5. In right triangle \(AMO_1\), \(MO_1=\sqrt{5^2-4^2}=3\,\text{cm}\); similarly, \(MO_2=3\,\text{cm}\).

Draw radius-\(5\,\text{cm}\) circles centered at \(A\) and \(B\). Their two intersections are the two possible centers. Each is \(3\,\text{cm}\) from the midpoint of \(\overline{AB}\).

- Construct the base perpendicular bisector from its two endpoints, not from the vertex. - Use \(AB=AC\) as an equal-distance statement about point \(A\). - For the angle-bisector claim, compare the two right triangles on either side of the constructed line.

1. Draw equal-radius arcs centered at \(B\) and \(C\), using a radius greater than half of \(BC\). Draw the line through the two arc intersections; this is the perpendicular bisector of \(BC\). 2. Since \(AB=AC\), point \(A\) is equidistant from \(B\) and \(C\). Therefore, \(A\) lies on the perpendicular-bisector locus, so the constructed line passes through \(A\). 3. The line meets \(BC\) at its midpoint and is perpendicular to \(BC\), so from vertex \(A\) it is both a median and an altitude. 4. Let the midpoint be \(M\). Right triangles \(ABM\) and \(ACM\) have congruent hypotenuses \(AB=AC\) and shared leg \(AM\), so they are congruent by HL. 5. Hence \(\angle BAM=\angle MAC\), so the line is also the angle bisector from \(A\).

a) Construct the perpendicular bisector of \(\overline{BC}\) with equal-radius arcs from \(B\) and \(C\). b) It passes through \(A\) because \(AB=AC\) makes \(A\) equidistant from \(B\) and \(C\). c) It is the median, altitude, and angle bisector from \(A\). The angle-bisector role follows from HL congruence of the two right triangles it forms.

- Use a property that every rhombus inherits from being a parallelogram. - Compare the two triangles that share \(\overline{AM}\). - After proving the angles at \(M\) are congruent, use their linear-pair relationship.

1. A rhombus is a parallelogram, so its diagonals bisect each other. Therefore, \(BM=DM\). 2. In triangles \(ABM\) and \(ADM\), \(AB=AD\) because all sides of a rhombus are congruent, \(BM=DM\), and \(AM\) is shared. 3. Thus \(\triangle ABM\cong\triangle ADM\) by SSS, so \(\angle BMA\cong\angle AMD\). 4. These two angles form a linear pair. Congruent supplementary angles each measure \(90^\circ\), so \(AC\perp BD\). 5. Since \(M\) is the midpoint of \(\overline{BD}\), line \(AC\) is the perpendicular bisector of \(\overline{BD}\).

The diagonals of the rhombus bisect each other, so \(BM=DM\). Triangles \(ABM\) and \(ADM\) are congruent by SSS, making the adjacent angles at \(M\) congruent; because they form a linear pair, both are right angles. Therefore, \(\overline{AC}\) is the perpendicular bisector of \(\overline{BD}\).

- Begin with what reflection across the perpendicular bisector of \(\overline{AB}\) does to \(A\) and \(B\). - Use the congruent base angles and congruent legs to track the images of the nonbase sides. - If a reflection exchanges two points, what is its mirror line relative to the segment joining them?

1. The base angles at \(A\) and \(B\) of an isosceles trapezoid are congruent. 2. Reflection across the perpendicular bisector \(s\) exchanges \(A\) and \(B\), while preserving angle measure and distance. 3. The image of ray \(\overrightarrow{AD}\) is ray \(\overrightarrow{BC}\). Since \(AD=BC\), the reflection maps \(D\) to \(C\). 4. Therefore, \(C\) and \(D\) are mirror images across \(s\), so \(s\) perpendicularly bisects \(\overline{CD}\).

Reflection across \(s\) exchanges \(A\) with \(B\) and, using the isosceles-trapezoid symmetry, exchanges \(D\) with \(C\). Thus \(s\) is the perpendicular bisector of \(\overline{CD}\).

- Identify the complete locus of points equidistant from \(A\) and \(B\). - Intersect that locus with the given circle. - Use the perpendicular from the circle's center to the resulting chord.

1. Every point equidistant from \(A\) and \(B\) lies on the perpendicular bisector \(p\) of \(\overline{AB}\). 2. Therefore, the required points are the intersections \(P\) and \(Q\) of \(p\) with the circle. 3. Let \(M\) be the foot of the perpendicular from \(O\) to \(p\). Since \(OM=3\,\text{cm}\), the perpendicular from the center to chord \(\overline{PQ}\) bisects the chord. 4. In right triangle \(OMP\), \(MP=\sqrt{5^2-3^2}=4\,\text{cm}\). 5. Thus \(PQ=2MP=8\,\text{cm}\).

The points are the two intersections \(P\) and \(Q\) of the circle with the perpendicular bisector of \(\overline{AB}\). Their distance is \(8\,\text{cm}\).

- Use the equal radii from each center to compare distances to \(C\) and \(D\). - Introduce one variable for the two parts of \(\overline{AB}\). - Compare the two right-triangle equations before deciding whether \(X\) is a midpoint.

1. Since \(AC=AD=8\,\text{cm}\), point \(A\) lies on the perpendicular bisector of \(\overline{CD}\). Since \(BC=BD=6\,\text{cm}\), point \(B\) lies on the same perpendicular bisector. 2. Therefore, line \(AB\) is the perpendicular bisector of \(\overline{CD}\), so \(CD\perp AB\). The perpendicular part of Chisom’s claim is true. 3. Let \(AX=x\), so \(XB=10-x\). The perpendicular line creates right triangles \(AXC\) and \(BXC\). 4. Thus \(x^2+CX^2=8^2\) and \((10-x)^2+CX^2=6^2\). 5. Subtracting gives \(x^2-(10-x)^2=28\), so \(20x-100=28\) and \(x=6.4\). 6. Therefore, \(AX=6.4\,\text{cm}\) and \(XB=3.6\,\text{cm}\), so \(X\) is not the midpoint of \(AB\). 7. If the two construction circles had equal radii, both \(C\) and \(D\) would be equidistant from \(A\) and \(B\), making \(CD\) the perpendicular bisector of \(AB\).

\(CD\perp AB\), but it does not bisect \(AB\). The lengths are \(AX=6.4\,\text{cm}\) and \(XB=3.6\,\text{cm}\). Equal circle radii would make \(CD\) the perpendicular bisector of \(AB\).

- Express tangency to each line as a perpendicular-distance condition. - Recall the locus of points equidistant from two intersecting lines. - Compare the halves of two adjacent supplementary angles.

1. A circle tangent to both \(\ell\) and \(m\) has a center whose perpendicular distances to the two lines are equal. 2. The locus of points equidistant from two intersecting lines consists of the bisectors of the four angles formed by the lines. 3. The app displays one full bisector line through the internal bisectors of a pair of vertical angles and a second full bisector line through the other pair. 4. Every point on either bisector line, except \(O\), can serve as a center; its perpendicular distance to either original line is the circle's positive radius. 5. Adjacent angles formed by \(\ell\) and \(m\) are supplementary. Their half-measures sum to \(90^\circ\), so the two bisector lines are perpendicular.

The complete locus is the union of the internal and external angle-bisector lines through \(O\), excluding \(O\) for circles with positive radius. The two locus lines are perpendicular because they bisect adjacent supplementary angles.

- Use the altitude relationships to identify the angles at \(E\) and \(F\). - Ask what single segment is subtended by both of those right angles. - Once you identify a diameter, determine the circle's center from it.

1. Since \(A\), \(E\), and \(C\) are collinear while \(E\) and \(H\) lie on the altitude from \(B\), \(AE\perp EH\). Thus \(\angle AEH=90^\circ\). 2. Since \(A\), \(F\), and \(B\) are collinear while \(F\) and \(H\) lie on the altitude from \(C\), \(AF\perp FH\). Thus \(\angle AFH=90^\circ\). 3. Points that form a right angle with endpoints \(A\) and \(H\) lie on the circle with diameter \(\overline{AH}\). Therefore, both \(E\) and \(F\) lie on that circle. 4. Construct the midpoint of \(\overline{AH}\) and draw the circle centered there through \(A\). It also passes through \(E\), \(H\), and \(F\).

Angles \(AEH\) and \(AFH\) are right angles, so \(E\) and \(F\) lie on the circle with diameter \(\overline{AH}\). Construct that circle from the midpoint of \(\overline{AH}\).

- Convert each condition into a locus before looking for the points. - One locus is built from a perpendicular-bisector construction; the other uses the midpoint of \(\overline{AB}\) as a circle center. - The two conditions make the resulting triangle both right and isosceles.

1. Construct the perpendicular bisector of \(\overline{AB}\) with equal-radius arcs from \(A\) and \(B\). This is the locus \(PA=PB\). 2. Construct the midpoint \(M\) of \(\overline{AB}\), then draw the circle centered at \(M\) through \(A\) and \(B\). By the diameter theorem, this is the locus of points \(P\ne A,B\) for which \(\angle APB=90^\circ\). 3. The two loci meet at exactly two points, one on each side of \(\overline{AB}\); those are all solutions. 4. For either solution, triangle \(APB\) is right isosceles with hypotenuse \(12\,\text{cm}\). If \(PA=PB=x\), then \(2x^2=12^2\). 5. Thus \(x=6\sqrt2\,\text{cm}\).

Construct the perpendicular bisector of \(\overline{AB}\) and the circle with diameter \(\overline{AB}\). Their two intersections are exactly the required points. For either one, \(PA=6\sqrt2\,\text{cm}\).

- A non-tangent choice of \(A\) gives a second circle-line intersection. - Identify the diameter and the inscribed angle that subtends it. - Compare the diameter method with the standard equal-arc method.

1. At most one choice of \(A\) makes the circle tangent to \(\ell\) at \(A\), namely the foot of the perpendicular from \(P\). Choosing a different point guarantees a second intersection \(H\ne A\). 2. Segment \(\overline{AP}\) is a diameter of the circle because \(M\) is its midpoint and the circle passes through \(A\) and \(P\). 3. Point \(H\) lies on the circle, so \(\angle AHP\) intercepts diameter \(\overline{AP}\). 4. An inscribed angle that intercepts a diameter is a right angle, so \(\angle AHP=90^\circ\). 5. Points \(A\) and \(H\) lie on \(\ell\), so \(\overline{PH}\perp\ell\). 6. This method creates the right angle using a circle with diameter \(\overline{AP}\), rather than using two intersecting equal-radius arcs.

Since \(\overline{AP}\) is a diameter, \(\angle AHP=90^\circ\). Because \(A\) and \(H\) lie on \(\ell\), \(\overline{PH}\perp\ell\). This method uses the diameter-right-angle theorem instead of equal-radius arcs.

- Replace one leg of the broken path with an equal reflected segment. - After reflection, compare a broken line with a straight segment. - Equality in the triangle inequality identifies the minimizing point.

1. Reflection preserves distance, so for every point \(X\) on \(\ell\), \(XB=XB'\). 2. Therefore, \(AX+XB=AX+XB'\). 3. By the triangle inequality, \(AX+XB'\ge AB'\), with equality exactly when \(A\), \(X\), and \(B'\) are collinear. 4. The app-created point \(P\) lies on both \(\ell\) and line \(AB'\), so \(AP+PB=AP+PB'=AB'\), the least possible value.

The reflection gives \(PB=PB'\), so \(AP+PB=AP+PB'=AB'\). For any other point \(X\) on \(\ell\), the triangle inequality gives \(AX+XB=AX+XB'\ge AB'\). Therefore, \(P\) is the minimizing point.

- Reverse the theorem about the segment from a circle's center to a chord midpoint. - Determine the direction any chord with midpoint \(M\) would have to take. - Use uniqueness of a perpendicular through a point for the uniqueness argument.

1. Construct the line \(m\) through \(M\) perpendicular to \(\overline{OM}\), and label its circle intersections \(A\) and \(B\). 2. Since \(OM\perp AB\), the perpendicular from the center to chord \(\overline{AB}\) bisects the chord. Thus \(AM=MB\). 3. If another chord had midpoint \(M\), the line from \(O\) to that midpoint would also be perpendicular to the chord. 4. There is only one line through \(M\) perpendicular to \(OM\), so any such chord must lie on \(m\) and must be \(\overline{AB}\).

Draw the perpendicular to \(OM\) through \(M\); its intersections with the circle are the chord endpoints. The chord is unique because only one line through \(M\) is perpendicular to \(OM\).

- One standard construction can give both the midpoint of \(\overline{AC}\) and a perpendicular through it. - The two missing vertices must be the same distance from \(M\) as \(A\) and \(C\). - Compare the two diagonals after the vertices have been constructed.

1. Construct the perpendicular bisector of \(\overline{AC}\); its intersection with \(\overline{AC}\) is the midpoint \(M\), and the same constructed line is perpendicular to \(AC\). 2. Set the compass opening to \(AM\). With center \(M\), draw a circle. Its two intersections with the perpendicular line are \(B\) and \(D\). 3. Then \(AM=CM=BM=DM=5\,\text{cm}\), so the diagonals \(AC\) and \(BD\) bisect each other, are congruent, and are perpendicular. 4. A quadrilateral whose diagonals have those three properties is a square. 5. Right triangle \(AMB\) has legs \(5\,\text{cm}\) and \(5\,\text{cm}\), so \(AB=5\sqrt2\,\text{cm}\).

Construct the perpendicular bisector of \(\overline{AC}\) to obtain midpoint \(M\). Draw the circle centered at \(M\) with radius \(AM\); its two intersections with the perpendicular bisector are \(B\) and \(D\). The resulting quadrilateral is a square, with side length \(5\sqrt2\,\text{cm}\).

- Represent each equidistance condition as a locus. - The locus equidistant from two parallel lines is the parallel line halfway between them. - Use the stated nonparallel condition to justify that the two loci have exactly one intersection.

1. The app creates the perpendicular bisector \(p\) of \(\overline{AB}\). Every point on \(p\) is equidistant from \(A\) and \(B\). 2. The app creates a common perpendicular to \(\ell\) and \(m\), finds the midpoint of the segment between the two lines, and draws line \(q\) through that midpoint parallel to \(\ell\). 3. Line \(q\) is the locus of points equidistant from the parallel lines \(\ell\) and \(m\). 4. Let \(P=p\cap q\). Then \(P\) satisfies both equidistance conditions. 5. Because \(p\) is not parallel to \(\ell\), it is not parallel to \(q\). Thus \(p\) and \(q\) intersect at exactly one point, so \(P\) is unique.

The app intersects the perpendicular bisector of \(\overline{AB}\) with the line halfway between \(\ell\) and \(m\). Their unique intersection is the required point \(P\).