Internal angle bisector of angle A of a \triangle ABC meets its circumcircle at P. If PC is of length 3 units, then find the length PI \times PB. I is the incenter.
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Step 1: Let the internal angle bisector of angle A meet BC at D. Show more…
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Derek F.
you will derive expressions for the radii of the circumscribed circle and the inscribed circle for $\triangle A B C .$ In these exercises, assume as given the following two results from geometry: i. The three angle bisectors of the angles of a triangle meet in a point. This point (labeled I in the figure) is the center of the inscribed circle. ii. The perpendicular bisectors of the sides of a triangle meet in a point. This point (labeled O in the figure) is the center of the circumscribed circle. Let $r$ denote the radius of the inscribed circle for $\triangle A B C$ and (as in the previous exercise) let $\mathscr{A}$ denote the area of $\triangle A B C$. Follow parts (a) through (c) to show that $r=\frac{2 . d}{a+b+c}$ (a) In the following figure, the inscribed circle (with center $I$ ) is tangent to side $A C$ at the point $D$. According to a theorem from geometry, ID is perpendicular to $A C$. What theorem is this? (State the theorem in complete sentences.) Then explain why the area of $\triangle A I C$ is $\frac{1}{2} r b$ (b) In the figure accompanying part (a), draw line segments $\overline{I A}, \overline{I B},$ and $\overline{I C},$ and then explain why $\vec{A}=\frac{1}{2} r b+\frac{1}{2} r c+\frac{1}{2} r a$ (c) Solve the equation in part (b) for $r$. You should obtain $r=2.9 /(a+b+c),$ as required.
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