Inradii Ratio (Posted on 2014-03-08)

            r1     |AP| 
Prove that ---- = ------
            r2     |PC|

	Submitted by Bractals
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A couple of lemmas without proof: The area of a triangle is equal to its inradius times its semiperimeter. The distance from a vertex of a triangle to the nearest point of tangency between the triangle and its incircle is equal to its semiperimeter minus the length of the side opposite the vertex. In our problem let    a = |AD| = |BC|,    b = |AC|,    c = |AB| = |CD|,    d = |DP|, and    hd = distance from D to the diagonal AC. Let a(XYZ), r(XYZ), and s(XYZ) denote the area, inradius, and semiperimeter of ΔXYZ respectively. If s = s(ABC), then    |AP| = s-a  and  |PC| = s-c. Therefore, r1 r(APD) ---- = -------- r2 r(PCD) a(APD) s(PCD) = -------- * -------- s(APD) a(PCD) |AP|hd/2 (|PC|+|CD|+|DP|)/2 = -------------------- * -------------------- (|AP|+|AD|+|DP|)/2 |PC|hd/2 |AP|*([s-c] + c + d) = ---------------------- |PC|*([s-a] + a + d) |AP|*(s + d) = -------------- |PC|*(s + d) |AP| = ------ |PC| QED

	Subject	Author	Date
	Analytic Solution	Jer	2014-03-10 23:45:18