- 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 \(\overline{AB}\).

- 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 \(5.625^\circ\).

- 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 \(\overline{AB}\).

- 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.

- 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.

- Think about how to combine the two given angle measures using exact copies. - After combining them, what construction corresponds to dividing their sum by \(2\)? - Check that the final construction represents the arithmetic mean, not merely one of the original angles.

1. From a new vertex, copy the \(50^\circ\) angle. 2. Starting from the terminal ray of that copy, copy the \(80^\circ\) angle adjacently on the same side. The combined angle measures \(50^\circ+80^\circ=130^\circ\). 3. Bisect the \(130^\circ\) angle. 4. Each resulting angle measures \(\frac{130^\circ}{2}=65^\circ\), which equals \(\frac{50^\circ+80^\circ}{2}\).

Copy the two source angles adjacently to form a \(130^\circ\) angle, then bisect it. Each half measures \(65^\circ\), the average of \(50^\circ\) and \(80^\circ\).

- 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.

- 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 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 \(AB\).

- 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, \(PA=6\sqrt2\,\text{cm}\).

- 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 \(5\sqrt2\,\text{cm}\).