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Geometry Difficulty 4.4 AIME Prove it South Africa

An altitude from $A$ to the triangle $ABC$ intersects the side $BC$ in $D$ . A circle touches $BC$ in $D$ , intersects $AB$ in $M$ and $N$ and intersects $AC$ in $P$ and $Q$ . Prove that
$\frac{AM + AN}{AC} = \frac{AP + AQ}{AB}.$

Solution

Let $P(X)$ denote the power of point $X$ with respect to the circle $M, N, D$ . Then
$\begin{align*} AB(AM + AN) &= AB \cdot AM + AB \cdot AN \\ &= (AN + NB)AM + AB(AB - BN) \\ &= AN \cdot AM + BN \cdot AM + AB^2 - AB \cdot BN \\ &= P(A) + AD^2 + BD^2 - BN(AB - AM) \\ &= P(A) + AD^2 + P(B) - BN \cdot BM \\ &= P(A) + AD^2 + P(B) - P(B) \\ &= P(A) + AD^2. \end{align*}$
A similar argument shows that $AC(AP + AQ) = P(A) + AD^2$ , which concludes the proof.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.