- Identify the arc intercepted by \(\angle ACB\). - How is an inscribed angle related to the measure of its intercepted arc? - Point \(C\) is on the major arc, so the angle intercepts the minor arc \(AB\).

1. Angle \(ACB\) is an inscribed angle that intercepts minor arc \(AB\). 2. An inscribed angle measures half its intercepted arc, so \(m\angle ACB=\frac{1}{2}\cdot86^\circ=43^\circ\).

\(43^\circ\)

- Check which arc the angle at \(C\) faces. - Compare an inscribed angle with the central angle facing the same arc.

1. Central angle \(AOB\) measures \(100^\circ\), so minor arc \(AB\) measures \(100^\circ\). 2. Inscribed angle \(ACB\) intercepts that arc and measures half as much. 3. Therefore, \(m\angle ACB=50^\circ\).

\(50^\circ\). An inscribed angle is half the measure of its intercepted arc.

- Trace the intercepted arc for each angle. - What determines an inscribed angle's measure?

1. Both vertices \(C\) and \(D\) lie on the other arc from \(A\) to \(B\). 2. Both inscribed angles intercept exactly the same minor arc \(AB\). 3. Each angle equals half that arc's measure, so they are equal.

Neither is larger: \(\angle ACB=\angle ADB\). Both angles intercept the same arc \(AB\), so each measures half that arc.

- Which angles of a cyclic quadrilateral are paired? - What is their total measure?

1. Opposite angles of a quadrilateral inscribed in a circle are supplementary. 2. The marked angle at \(B\) is \(110^\circ\), so the angle at \(D\) is \(180^\circ-110^\circ=70^\circ\).

\(70^\circ\)

- Identify the arc intercepted by both marked angles. - What is the relationship between a central angle and an inscribed angle that intercept the same arc?

1. The inscribed angle and the marked central angle intercept the same arc \(BC\). 2. An inscribed angle has half the measure of the central angle that intercepts the same arc, so \(\alpha=\frac{1}{2}\cdot112^\circ=56^\circ\).

\(\alpha=56^\circ\)

- Identify the arc intercepted by the marked inscribed angle. - Which is larger: the central angle or an inscribed angle intercepting the same arc?

1. The marked inscribed angle and central angle intercept the same arc \(PQ\). 2. A central angle has twice the measure of an inscribed angle that intercepts the same arc, so \(\mu=2\cdot34^\circ=68^\circ\).

\(\mu=68^\circ\)

- Which arc does the angle at \(S\) intercept? - How does an inscribed angle compare with its intercepted arc? - Which two sides meet at the right angle?

1. Angle \(\angle ASB\) intercepts the semicircle from \(A\) to \(B\), whose arc measure is \(180^\circ\). 2. An inscribed angle measures half its intercepted arc, so \(\gamma=90^\circ\). 3. The legs of the right triangle are \(\overline{AS}\) and \(\overline{BS}\).

\(\gamma=90^\circ\); the legs are \(\overline{AS}\) and \(\overline{BS}\).

- Find the angle that subtends the diameter. - Use the angle marked in the diagram with the triangle angle sum.

1. Since \(\angle ACB\) subtends diameter \(\overline{AB}\), \(\angle ACB=90^\circ\). 2. Use the triangle angle sum: \(\beta=180^\circ-90^\circ-28^\circ=62^\circ\).

\(\beta=62^\circ\)

- What special relationship holds between opposite angles of a quadrilateral inscribed in a circle? - Combine that relationship with the angle measure shown in the diagram.

1. Opposite angles of a cyclic quadrilateral are supplementary. 2. Therefore, \(\delta=180^\circ-104^\circ=76^\circ\).

\(\delta=76^\circ\)

- What is true of inscribed angles that intercept the same arc? - How are a central angle and an inscribed angle intercepting the same arc related? - Use the numerical angle marked in the diagram.

1. The diagram shows \(\alpha=40^\circ\). Inscribed angles that intercept the same arc are congruent, so \(\beta=\alpha=40^\circ\). 2. A central angle has twice the measure of an inscribed angle intercepting the same arc. 3. Therefore, \(\mu=2\cdot40^\circ=80^\circ\).

\(\beta=40^\circ\) and \(\mu=80^\circ\)

- Identify the arc intercepted by each marked angle. - What is true of inscribed angles that intercept the same arc? - Look for pairs of angles with the same arc endpoints. - Name the theorem involving angles whose vertices lie on a circle.

1. The diagram shows \(m\angle CAD=75^\circ\). Angles \(\angle CAD\) and \(\alpha=\angle CBD\) are inscribed angles that intercept the same arc \(CD\). 2. Inscribed angles intercepting the same arc are congruent, so \(\alpha=75^\circ\). 3. The diagram also shows \(m\angle ABD=35^\circ\). Angles \(\angle ABD\) and \(\beta=\angle ACD\) intercept the same arc \(AD\). 4. Therefore, \(\beta=35^\circ\). 5. The theorem used is the Inscribed Angle Theorem, specifically the result that inscribed angles intercepting the same arc are congruent.

\(\alpha=75^\circ\) and \(\beta=35^\circ\). The theorem is the Inscribed Angle Theorem.

- Identify the arc intercepted by each angle of the quadrilateral. - Recall the angle measure subtended by a diameter. - Distinguish the conditions for a rectangle from the stronger conditions for a square.

1. Angle \(\angle ABC\) intercepts diameter \(\overline{AC}\), so it is a right angle. 2. Angle \(\angle BCD\) intercepts diameter \(\overline{BD}\), so it is a right angle. 3. Similarly, \(\angle CDA\) intercepts diameter \(\overline{AC}\), and \(\angle DAB\) intercepts diameter \(\overline{BD}\). Both are right angles. 4. Therefore, \(ABCD\) has four right angles and is a rectangle. 5. Perpendicular diameters are not required; they would only make the rectangle a square.

Each vertex angle intercepts one of the two diameters and is therefore \(90^\circ\). Thus \(ABCD\) is always a rectangle. It is a square only when the diameters are perpendicular.

- What does a diameter tell you about an inscribed angle that subtends it? - Identify the hypotenuse before choosing a right-triangle relation. - Check that the unknown side must be shorter than the diameter.

1. Because \(\overline{AB}\) is a diameter and \(C\) lies on the circle, \(\angle ACB=90^\circ\). 2. In right triangle \(ACB\), \(AB=13\,\text{cm}\) is the hypotenuse and \(AC=5\,\text{cm}\) is one leg. 3. By the Pythagorean theorem, \(BC^2=13^2-5^2=144\). 4. Therefore, \(BC=12\,\text{cm}\).

\(12\,\text{cm}\)

- Does an inscribed angle intercept the arc containing its vertex? - Which path from \(A\) to \(B\) does the angle at \(C\) face? - How does an inscribed angle's measure relate to the arc it faces?

1. The minor arc \(AB\) measures \(140^\circ\), matching the central angle. 2. Because \(C\) lies on that minor arc, \(\angle ACB\) intercepts the other arc, measuring \(360^\circ-140^\circ=220^\circ\). 3. The inscribed angle is half of \(220^\circ\), or \(110^\circ\).

\(110^\circ\). The inscribed angle intercepts the major arc \(AB\), not the arc containing \(C\).

- Extend attention from \(\angle APC\) to its vertical angle. - Which arc faces each of those angles? - How does the intersecting-chords theorem combine their measures?

1. The angle formed by two intersecting chords inside a circle is half the sum of the two arcs intercepted by the angle and its vertical angle. 2. Those arcs are \(AC\) and \(BD\), so \(m\angle APC=\frac{90^\circ+100^\circ}{2}=95^\circ\).

\(95^\circ\). The angle and its vertical angle intercept the two arcs whose measures are averaged.

- Compare where the angle's vertex sits with the circle. - Identify the arc between the far intersection points and the arc between the near points. - Which operation distinguishes an exterior angle from the sum rule for intersecting chords inside?

1. For two secants from a point outside a circle, the exterior angle is half the difference between the far and near intercepted arcs. 2. Thus the angle at \(P\) measures \(\frac{110^\circ-50^\circ}{2}=30^\circ\).

\(30^\circ\), using \(\frac{110^\circ-50^\circ}{2}\).

- Mark the point where the tangent touches the circle. - Which arc reaches the farther secant intersection, and which reaches the nearer one? - Compare the rule with the two-secant exterior angle rule.

1. The exterior angle between a tangent and secant is half the difference of the intercepted arcs. 2. The larger arc from \(B\) to \(T\) measures \(120^\circ\); the smaller arc from \(A\) to \(T\) measures \(60^\circ\). 3. Thus \(m\angle TPA=\frac{120^\circ-60^\circ}{2}=30^\circ\).

\(30^\circ\), using half the difference between arcs \(BT\) and \(AT\).

- What is special about opposite angles when all vertices lie on a circle? - Use one equation for the two expressions. - Check the resulting measures against their required total.

1. Opposite angles in a cyclic quadrilateral total \(180^\circ\). 2. Hence \((3x+10)+(2x+20)=180\), or \(5x+30=180\), so \(x=30\). 3. The angles are \(3(30)+10=100^\circ\) and \(2(30)+20=80^\circ\).

\(x=30\). The opposite angles measure \(100^\circ\) and \(80^\circ\). They sum to \(180^\circ\) because opposite angles of a quadrilateral inscribed in a circle are supplementary.

- Where is the center when a segment is a diameter? - Which circle points give a right angle at \(C\), and where must \(C\) be for the two legs to match? - How can you describe all points a fixed perpendicular distance from a line?

1. The center is the midpoint \(M\) of \(\overline{AB}\). Its radius is \(3\,\text{cm}\). 2. Every point on this circle except \(A\) and \(B\) makes a right angle at \(C\) because \(\overline{AB}\) is a diameter. 3. For \(AC=BC\), point \(C\) must also lie on the perpendicular bisector of \(\overline{AB}\). That line meets the circle in two points. 4. Each such point is \(3\,\text{cm}\) from \(\overline{AB}\), so the altitude is \(3\,\text{cm}\). 5. Points \(2\,\text{cm}\) from line \(AB\) lie on two parallel lines, one on each side. Each crosses the circle twice, giving four choices of \(D\).

a) Draw a circle centered at the midpoint of \(\overline{AB}\) with radius \(3\,\text{cm}\). b) The two intersections with the perpendicular bisector of \(\overline{AB}\); altitude \(3\,\text{cm}\). c) Four points. The two lines parallel to \(AB\) and \(2\,\text{cm}\) away, one on each side, each meet the circle twice.

- Every possible right-angle vertex lies on the circle with the hypotenuse as diameter. - How can the given angle at \(A\) select the correct point on that circle? - The farthest point on the circle from the diameter is one radius away. - Use \(A=\frac{1}{2}bh\) with the hypotenuse as the base.

1. Construct the midpoint \(M\) of \(\overline{AB}\) and the circle centered at \(M\) with radius \(4\,\text{cm}\). 2. From \(A\), construct the two rays that make a \(40^\circ\) angle with \(\overline{AB}\), one on each side of the line. Each ray has a second intersection with the circle, giving the two mirror-image possible locations of \(C\). Because \(\overline{AB}\) is a diameter, \(\angle C=90^\circ\) in each triangle. 3. Any possible right-angle vertex \(C\) lies on this circle. The altitude \(h_c\) is the perpendicular distance from \(C\) to \(\overline{AB}\), and the greatest such distance is the circle's radius, \(4\,\text{cm}\). 4. The maximum area is \(\frac{1}{2}\cdot8\,\text{cm}\cdot4\,\text{cm}=16\,\text{cm}^2\). 5. Since \(20\,\text{cm}^2>16\,\text{cm}^2\), such a triangle is impossible.

a) Construct the circle with diameter \(\overline{AB}\). The two possible points \(C\) are the second intersections of that circle with the two rays from \(A\) that make \(40^\circ\) with \(\overline{AB}\), one on each side. b) The maximum altitude is the radius, \(4\,\text{cm}\). c) No. The maximum possible area is \(16\,\text{cm}^2\).

- Recall the measure of an inscribed angle that subtends a diameter. - Use the triangle angle sum after identifying the right angle. - Compare an angle subtending the same segment from a point on the circle and from a point inside it.

1. An inscribed angle that subtends a diameter is a right angle, so \(\gamma=90^\circ\). 2. By the triangle angle-sum theorem, \(\alpha=180^\circ-90^\circ-38^\circ=52^\circ\). 3. When \(C\) is moved inside the circle, the angle subtending fixed segment \(\overline{AB}\) becomes greater than \(90^\circ\). Therefore, \(\gamma\) becomes obtuse.

a) \(\gamma=90^\circ\), by the inscribed-angle theorem for a diameter. b) \(\alpha=52^\circ\). c) \(\gamma>90^\circ\), so the angle becomes obtuse.

- Which circle makes a given segment the hypotenuse of a right triangle? - Where must the vertex of an isosceles triangle lie relative to its base? - What is the relationship between a circle's diameter and radius?

1. Every point \(C\) on the circle with diameter \(\overline{AB}\), other than \(A\) and \(B\), gives a right angle at \(C\). The center is midpoint \(M\), and the radius is \(4\,\text{cm}\). 2. For \(AC=BC\), point \(C\) lies on the perpendicular bisector of \(\overline{AB}\). Its two intersections with the circle are each \(4\,\text{cm}\) from line \(AB\), so the altitude is \(4\,\text{cm}\). 3. Since \(C\) lies on the circle centered at \(M\), \(MC\) is a radius and \(AB\) is a diameter. Thus \(MC=\frac{1}{2}AB\).

a) Every point on the circle with diameter \(\overline{AB}\) except \(A\) and \(B\); its radius is \(4\,\text{cm}\). b) \(4\,\text{cm}\) c) \(MC\) is a radius of the circle whose diameter is \(AB\), so \(MC=\frac{1}{2}AB\).

- An inscribed angle that subtends a diameter is a right angle. - A point reflection across the center sends a point on the circle to the opposite endpoint of a diameter. - Examine the diagonals and interior angles of the resulting quadrilateral.

1. Since \(\overline{AB}\) is a diameter and \(C\) lies on the circle, \(\angle ACB=90^\circ\). 2. The third angle is \(\angle ABC=180^\circ-90^\circ-60^\circ=30^\circ\). 3. Reflecting \(C\) across \(M\) makes \(\overline{CD}\) another diameter. Thus, diagonals \(\overline{AB}\) and \(\overline{CD}\) of quadrilateral \(ACBD\) bisect each other at \(M\) and are congruent. 4. Each interior angle of \(ACBD\) subtends one of the diameters and therefore measures \(90^\circ\). Hence, \(ACBD\) is a rectangle.

a) \(\angle ACB=90^\circ\) and \(\angle ABC=30^\circ\). b) \(ACBD\) is a rectangle because its diagonals are congruent diameters that bisect each other, and each interior angle is a right angle.

- First determine the circle whose diameter is \(\overline{AB}\). - How does the position of \(P\) relative to that circle determine whether \(\angle APB\) is acute, right, or obtuse? - Compare each listed position with the circle without calculating the angle directly.

1. The circle with diameter \(\overline{AB}\) has center \((3,1)\) and radius \(2\). 2. Point \((3,1.5)\) is \(0.5\) unit from the center, so it is inside the circle and \(\angle APB\) is obtuse. 3. Point \((3,3)\) is \(2\) units from the center, so it lies on the circle and \(\angle APB\) is right. 4. Point \((3,5)\) is \(4\) units from the center, so it is outside the circle and \(\angle APB\) is acute. 5. Therefore, the angle decreases from obtuse to right to acute as \(P\) moves upward. The right angle occurs at \(P(3,3)\).

a) At \(P(3,1.5)\), the angle is obtuse; at \(P(3,3)\), it is right; and at \(P(3,5)\), it is acute. The angle decreases as \(P\) moves upward. b) \(P(3, 3)\)

- Start with the midpoint and length of \(\overline{AB}\). - Compare the distance from the circle's center to the horizontal walking path with the radius. - What angle is subtended by a diameter from a point on its circle? - How do points outside that diameter circle classify the angle subtending \(\overline{AB}\)?

1. The midpoint of \(A(2,2)\) and \(B(10,2)\) is \((6,2)\), and \(AB=8\), so the circle with diameter \(\overline{AB}\) has center \((6,2)\) and radius \(4\). 2. The walking path lies on \(y=6\). Its distance from the center \((6,2)\) is \(4\), equal to the radius, so the line is tangent to the circle at \(P(6,6)\). 3. At \(P(6,6)\), point \(P\) lies on the circle with diameter \(\overline{AB}\), so \(\angle APB=90^\circ\). 4. Every other point of the walking segment lies outside that circle, so the angle subtending \(\overline{AB}\) is acute there. Therefore, the greatest viewing angle occurs at \(P(6,6)\) and measures \(90^\circ\).

a) Center: \((6,2)\); radius: \(4\) b) The path \(y=6\) is tangent to the circle at \(P(6,6)\). c) At the tangency point, \(P\) lies on the circle with diameter \(\overline{AB}\), so \(\angle APB\) is a right angle. All other positions on the path are outside the circle and give acute angles. The maximum is \(90^\circ\) at \(P(6,6)\).

- Use the diagram to identify which intercepted arc joins the farther points and which joins the nearer points. - Recall how an exterior angle formed by two secants is related to those two arc measures. - Check that your angle is smaller than either intercepted arc.

1. The measure of an angle formed by two secants outside a circle equals half the positive difference of the intercepted arcs. 2. Therefore, \(m\angle P=\frac{1}{2}(116^\circ-34^\circ)=\frac{1}{2}(82^\circ)=41^\circ\).

\(41^\circ\)

- Use the diagram to distinguish the tangent point from the two secant intersection points. - Recall how an exterior tangent-secant angle is related to the difference of the intercepted arcs. - Represent the unknown larger arc with a variable before solving.

1. For an angle formed by a tangent and a secant outside a circle, the angle measure is half the positive difference of the intercepted arcs. 2. Let \(x\) be the measure of the larger arc. Then \(35^\circ=\frac{1}{2}(x-42^\circ)\). 3. Multiply by \(2\): \(70^\circ=x-42^\circ\). 4. Therefore, \(x=112^\circ\).

\(112^\circ\)

- Which arc does \(\angle ABC\) intercept? - What theorem connects an inscribed angle with a diameter? - After establishing the right angle, determine the diameter and use the labeled side in the diagram. - How can the two perpendicular legs be used to find the area?

1. Because \(AC\) is a diameter and \(B\) lies on the circle, \(\angle ABC\) intercepts a semicircle. Therefore, \(\angle ABC=90^\circ\). 2. The diameter has length \(AC=2\cdot5=10\,\text{cm}\). 3. The diagram gives \(BC=6\,\text{cm}\). In right triangle \(ABC\), \(AB^2=10^2-6^2=64\), so \(AB=8\,\text{cm}\). 4. The area is \(\frac{1}{2}\cdot6\cdot8=24\,\text{cm}^2\).

a) \(\angle ABC=90^\circ\) because an inscribed angle that intercepts a diameter is a right angle. b) \(24\,\text{cm}^2\)

- What is special about an inscribed angle that intercepts a diameter? - Work separately in each of the two triangles sharing \(\overline{BD}\). - How can the two portions of the straight segment \(\overline{AC}\) be combined?

1. Each angle at \(D\) intercepts a diameter: \(\angle ADB=90^\circ\) and \(\angle BDC=90^\circ\). 2. In right triangle \(ABD\), \(AD=\sqrt{10^2-8^2}=6\,\text{cm}\). 3. In right triangle \(BCD\), \(DC=\sqrt{17^2-8^2}=15\,\text{cm}\). 4. Since \(D\) lies between \(A\) and \(C\), \(AC=AD+DC=21\,\text{cm}\).

a) Both are right angles because each subtends a diameter: \(\angle ADB=\angle BDC=90^\circ\). b) \(AC=21\,\text{cm}\)

- What do the matching angle marks at \(A\) tell you? - Replace \(\angle ABM\) with the corresponding angle that uses point \(C\) on the same line. - Which two inscribed angles intercept arc \(AC\)?

1. The matching angle marks at \(A\) show that \(\angle BAM\cong\angle KAC\). 2. Because \(M\) lies on \(BC\), \(\angle ABM=\angle ABC\). 3. Angles \(\angle ABC\) and \(\angle AKC\) are inscribed angles that intercept the same arc \(AC\), so they are congruent. 4. Therefore, \(\triangle ABM\sim\triangle AKC\) by AA.

\(\triangle ABM\sim\triangle AKC\) by AA because the angle bisector gives one congruent angle pair and the inscribed angle theorem gives \(\angle ABM\cong\angle AKC\).

- What can you conclude about arcs subtended by equal chords? - Which inscribed angles intercept those two arcs? - Identify a transversal through \(A\) and \(C\).

1. Equal chords \(AD\) and \(BC\) intercept equal minor arcs. 2. Angles \(\angle ACD\) and \(\angle CAB\) intercept those arcs, so their measures are equal. 3. These are alternate interior angles for lines \(CD\) and \(AB\) cut by transversal \(AC\). 4. By the converse of the alternate interior angles theorem, \(\overline{CD}\parallel\overline{AB}\).

Since \(AD=BC\), the corresponding intercepted arcs and inscribed angles are equal. Thus \(\angle ACD=\angle CAB\), so \(CD\parallel AB\).

- Identify the arc intercepted by each of the two opposite inscribed angles. - How are the two arcs joining the same pair of endpoints related to a full circle? - Apply the Inscribed Angle Theorem to each angle before adding them.

1. Inscribed angle \(\angle DAB\) intercepts arc \(DB\) that passes through \(C\), so \(m\angle DAB=\frac{1}{2}m\widehat{DCB}\). 2. Inscribed angle \(\angle BCD\) intercepts the other arc \(BD\), the one that passes through \(A\), so \(m\angle BCD=\frac{1}{2}m\widehat{BAD}\). 3. Those two intercepted arcs make the full circle, so \(m\widehat{DCB}+m\widehat{BAD}=360^\circ\). 4. Therefore, \(m\angle DAB+m\angle BCD=\frac{1}{2}(360^\circ)=180^\circ\).

Each opposite inscribed angle is half of one of the two arcs joining \(B\) and \(D\). Those arcs total \(360^\circ\), so the two angle measures total \(180^\circ\). Thus opposite angles of a cyclic quadrilateral are supplementary.

- What two sides in each triangle are radii? - Express each small angle at \(C\) using its triangle's center angle. - Add the small angles, then compare with the central angle around the other side of \(O\).

1. Put \(\alpha=m\angle AOC\) and \(\beta=m\angle BOC\). 2. Since \(OA=OC\), triangle \(AOC\) is isosceles and \(m\angle ACO=\frac{180^\circ-\alpha}{2}\). 3. Similarly, \(m\angle OCB=\frac{180^\circ-\beta}{2}\). 4. Adding gives \(m\angle ACB=180^\circ-\frac{\alpha+\beta}{2}\). 5. The other central angle \(AOB\), which faces the arc not containing \(C\), has measure \(360^\circ-\alpha-\beta\). Half of this is exactly \(180^\circ-\frac{\alpha+\beta}{2}\).

Isosceles triangle base angles give \(m\angle ACB=\frac{180^\circ-\alpha}{2}+\frac{180^\circ-\beta}{2}=\frac{360^\circ-\alpha-\beta}{2}=\frac{1}{2}m\angle AOB\) for the central angle facing the intercepted arc.

- Can you use the triangle formed by the crossing chords and the short chord joining two of their endpoints? - Which arcs face the angles at its two vertices on the circle? - How do the four arcs between consecutive points total a full turn?

1. In triangle \(APC\), angle \(PAC\) intercepts arc \(BC\), and angle \(ACP\) intercepts arc \(AD\). 2. Their measures are \(\frac{1}{2}m\widehat{BC}\) and \(\frac{1}{2}m\widehat{AD}\). 3. The triangle angle sum gives \(m\angle APC=180^\circ-\frac{1}{2}(m\widehat{BC}+m\widehat{AD})\). 4. The four arcs \(AC,CB,BD,DA\) total \(360^\circ\). Thus \(m\widehat{BC}+m\widehat{AD}=360^\circ-(m\widehat{AC}+m\widehat{BD})\). 5. Substitute to get \(m\angle APC=\frac{1}{2}(m\widehat{AC}+m\widehat{BD})\).

In triangle \(APC\), the angles at \(A\) and \(C\) are half of arcs \(BC\) and \(AD\). Their sum is \(\frac{1}{2}(360^\circ-m\widehat{AC}-m\widehat{BD})\). Subtracting this from \(180^\circ\) leaves \(m\angle APC=\frac{1}{2}(m\widehat{AC}+m\widehat{BD})\).

- Draw a chord to make a triangle with vertex \(P\). - One of that triangle's angles lies on the circle; another is supplementary to an inscribed angle. - Which arc does each inscribed angle intercept?

1. Join \(B\) to \(C\) to form triangle \(PBC\). 2. Angle \(PBC\) is an inscribed angle intercepting arc \(AC\), so its measure is \(\frac{1}{2}m\widehat{AC}\). 3. Angle \(PCB\) is supplementary to inscribed angle \(DCB\), so it measures \(180^\circ-\frac{1}{2}m\widehat{BD}\). 4. By the triangle angle sum, the angle at \(P\) is \(180^\circ-\frac{1}{2}m\widehat{AC}-(180^\circ-\frac{1}{2}m\widehat{BD})\). 5. Simplify to obtain \(m\angle P=\frac{1}{2}(m\widehat{BD}-m\widehat{AC})\).

In triangle \(PBC\), \(m\angle PBC=\frac{1}{2}m\widehat{AC}\) and \(m\angle PCB=180^\circ-\frac{1}{2}m\widehat{BD}\). Subtracting both from \(180^\circ\) gives \(m\angle P=\frac{1}{2}(m\widehat{BD}-m\widehat{AC})\).

- Which chord can you add to make a triangle containing the angle at \(P\)? - What arc does the angle at the far secant point intercept? - At the tangent point, one angle is supplementary to the tangent-chord angle you can find from an arc.

1. Join \(B\) to \(T\) and consider triangle \(PBT\). 2. Angle \(PBT\) intercepts arc \(AT\), so \(m\angle PBT=\frac{1}{2}m\widehat{AT}\). 3. At \(T\), the tangent-chord theorem shows that \(\angle PTB\) is supplementary to the angle measuring half of arc \(BT\). Hence \(m\angle PTB=180^\circ-\frac{1}{2}m\widehat{BT}\). 4. The angle sum of triangle \(PBT\) gives \(m\angle P=180^\circ-\frac{1}{2}m\widehat{AT}-(180^\circ-\frac{1}{2}m\widehat{BT})\). 5. Therefore \(m\angle P=\frac{1}{2}(m\widehat{BT}-m\widehat{AT})\).

Triangle \(PBT\) has \(m\angle PBT=\frac{1}{2}m\widehat{AT}\) and \(m\angle PTB=180^\circ-\frac{1}{2}m\widehat{BT}\). Subtract their sum from \(180^\circ\) to get \(m\angle P=\frac{1}{2}(m\widehat{BT}-m\widehat{AT})\).

- A point on the circle with diameter \(\overline{AB}\) forms a right angle subtending \(\overline{AB}\). - Points inside that circle form obtuse angles, and points outside it form acute angles. - Substitute \(y=3\) into the circle equation to find the right-angle positions.

1. The circle with diameter \(\overline{AB}\) has center \((4,0)\), radius \(4\), and equation \((x-4)^2+y^2=16\). 2. For \(P_1(0,3)\), \((0-4)^2+3^2=25>16\), so \(P_1\) is outside the circle and \(\angle AP_1B\) is acute. 3. For \(P_2(4,3)\), \((4-4)^2+3^2=9<16\), so \(P_2\) is inside the circle and \(\angle AP_2B\) is obtuse. 4. For \(P_3(8,3)\), \((8-4)^2+3^2=25>16\), so \(P_3\) is outside the circle and \(\angle AP_3B\) is acute. 5. On the path \(y=3\), right-angle positions lie on the circle and satisfy \((x-4)^2+9=16\). Thus, \(x=4\pm\sqrt{7}\). 6. Both values lie between \(0\) and \(8\), so the two positions are \(\left(4-\sqrt{7},3\right)\) and \(\left(4+\sqrt{7},3\right)\).

a) \(P_1\): acute \(P_2\): obtuse \(P_3\): acute b) The right-angle positions are \(\left(4-\sqrt{7},3\right)\) and \(\left(4+\sqrt{7},3\right)\).

- Use the midpoint and \(AB=2CD\) to compare \(AM\), \(MB\), and \(CD\). - Combine those relationships with \(DM=CD\). What does that tell you about the distances from \(M\) to \(A\), \(B\), and \(D\)? - Once you have identified the circle, what special role does \(AB\) have? - Only after proving the right angle should you use the dimensions in the diagram to find \(BD\).

1. The diagram shows \(CD=7.5\,\text{cm}\). Since \(AB=2CD\) and \(M\) is the midpoint of \(AB\), \(AM=MB=CD=7.5\,\text{cm}\). Because \(DM=CD\), \(DM=7.5\,\text{cm}\) as well. 2. Thus \(MA=MB=MD\), so \(A\), \(B\), and \(D\) lie on a circle centered at \(M\). Because \(M\) is the midpoint of \(AB\), segment \(AB\) is a diameter of that circle. Therefore, \(\angle ADB=90^\circ\). 3. The diameter is \(AB=15\,\text{cm}\). In right triangle \(ADB\), \(9^2+BD^2=15^2\). 4. Therefore, \(BD^2=144\), so \(BD=12\,\text{cm}\).

a) \(MA=MB=MD\), so \(A\), \(B\), and \(D\) lie on a circle centered at \(M\). Since \(AB\) is a diameter, \(\angle ADB=90^\circ\). b) \(BD=12\,\text{cm}\)

- Use \(AB=2CD\) and the midpoint definition to compare \(AM\), \(MB\), and \(CD\). - Since \(AM\parallel CD\) and \(AM=CD\), what can you conclude about quadrilateral \(AMCD\)? - Use \(AD=CD\) to compare \(MA\), \(MB\), and \(MC\). - Only after proving the right angle should you use the two dimensions shown in the diagram.

1. Let \(M\) be the midpoint of \(AB\). Since \(AB=2CD\), \(AM=MB=CD\). 2. Because \(AM\parallel CD\) and \(AM=CD\), quadrilateral \(AMCD\) is a parallelogram. Therefore, \(MC=AD\). 3. The problem gives \(AD=CD\), so \(MA=MB=MC\). Thus \(A\), \(B\), and \(C\) lie on a circle centered at \(M\), with \(AB\) as a diameter. Therefore, \(\angle ACB\) is a right angle. 4. From the diagram, \(AC=15\,\text{cm}\) and \(BC=20\,\text{cm}\). Apply the Pythagorean theorem: \(AB^2=15^2+20^2=625\), so \(AB=25\,\text{cm}\). 5. Since \(AB=2CD\), \(CD=12.5\,\text{cm}\).

a) \(MA=MB=MC\), so \(A\), \(B\), and \(C\) lie on a circle centered at \(M\). Since \(AB\) is a diameter, \(\angle ACB=90^\circ\). b) \(AB=25\,\text{cm}\) and \(CD=12.5\,\text{cm}\)

- What angle does diameter \(PQ\) subtend at \(M\)? What angle does diameter \(PS\) subtend there? - If both \(MQ\) and \(MS\) are perpendicular to the same line through \(M\), what does that tell you about \(Q\), \(M\), and \(S\)? - After proving that \(PM\) is an altitude of triangle \(PQS\), find \(QS\) from the side lengths shown. - Compare two base-height formulas for the area of triangle \(PQS\).

1. Since \(PQ\) is a diameter of the first circle, \(\angle PMQ=90^\circ\). Since \(PS\) is a diameter of the second circle, \(\angle PMS=90^\circ\). 2. Rays \(MQ\) and \(MS\) are therefore both perpendicular to \(MP\) at \(M\). They lie on the same line through \(M\), so \(Q\), \(M\), and \(S\) are collinear. Hence \(PM\perp QS\). 3. From the diagram, \(PQ=12\,\text{cm}\) and \(PS=5\,\text{cm}\), so \(QS=\sqrt{12^2+5^2}=13\,\text{cm}\). 4. Compute the area of right triangle \(PQS\) in two ways: \(\frac{1}{2}(12)(5)=\frac{1}{2}(13)(PM)\). 5. Thus \(60=13(PM)\), so \(PM=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\).

a) \(\angle PMQ=\angle PMS=90^\circ\) because they subtend diameters. Therefore \(Q\), \(M\), and \(S\) are collinear and \(PM\perp QS\). b) \(PM=\frac{60}{13}\,\text{cm}\approx4.62\,\text{cm}\)

- Which segments are perpendicular because they are altitudes? - What do the two right angles share as endpoints? - How do you construct the circle having a given segment as its diameter?

1. Since \(BE\perp AC\), segments \(AE\) and \(EH\) form a right angle at \(E\). 2. Since \(CF\perp AB\), segments \(AF\) and \(FH\) form a right angle at \(F\). 3. Both \(\angle AEH\) and \(\angle AFH\) are right angles intercepting \(\overline{AH}\). Thus \(E\) and \(F\) lie on the circle with diameter \(\overline{AH}\). 4. Find the midpoint of \(\overline{AH}\), then use it as the center and \(\frac{1}{2}AH\) as the radius.

Both \(E\) and \(F\) form right angles subtending \(\overline{AH}\). Construct the circle with diameter \(\overline{AH}\); it passes through all four points.