Triangle $ABC$ is right angled at $A$. The circle with centre $A$ and radius $AB$ cuts $BC$ and $AC$

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Triangle $ABC$ is right angled at $A$. The circle with centre $A$ and radius $AB$ cuts $BC$ and $AC$ internally at $D$ and $E$ respectively. If $BD = 20$ and $DC = 16$ then the length $AC$ equals:

Accepted Answer

[Key Concept: Right triangle with a circle centred at the right-angle vertex: use power of a point on the hypotenuse.]$(AC)^{2} + r^{2} = (36)^{2}$$$\begin{array}{r} (CE)(CF) = (CD)(BC)\#(1) \ \Rightarrow \ (AC - r)(AC + r) = 16 	imes 36\#(2) \end{array}$$On adding (1) and (2), we get$${\Rightarrow 2(AC)^{2} = 36 	imes 52 }{\Rightarrow AC = 6\sqrt{26} }$$

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