5.1.1 Perpendicular and Angle Bisector Theorems (continued)
By the Converse of the Angle Bisector Theorem, if a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle.
In this example, PB is found. It is given that ∠ BAP ≅ ∠ CAP . Therefore, AP bisects ∠ BAC , and so AP is an angle bisector of ∠ BAC . Since AP is an angle bisector, the Angle Bisector Theorem can be applied. The Angle Bisector Theorem states that if a point is on the bisector of an angle, then that point is equidistant from the sides of the angle. P is a point on angle bisector AP . Therefore, P must be equidistant from the sides of ∠ BAC . Since P must be equidistant from the sides of ∠ BAC , it follows that PC = PB . It is given that PC = 10. Thus, PB = 10 as well. In this example, P is equidistant from the sides of ∠ CAP since PB ≅ PC . Therefore, PA bisects ∠ BAC by the Converse of the Angle Bisector Theorem. Since PA bisects ∠ BAC , it follows that m ∠ BAP = m ∠ CAP . Expressions for m ∠ BAP and m ∠ CAP are given. Set these expressions equal to each other and solve for y . Then, substitute the value of y into the expression for m ∠ CAP to find m ∠ CAP .
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