# 2021 JMPSC Invitationals Problems/Problem 9

## Problem

In $\triangle ABC$, let $D$ be on $\overline{AB}$ such that $AD=DC$. If $\angle ADC=2\angle ABC$, $AD=13$, and $BC=10$, find $AC.$

## Solution 1

From the fact that $AD=DB$ and $\angle ADC = 2\angle ABC,$ we find that $\triangle ABC$ is a right triangle with a right angle at $C;$ thus by the Pythagorean Theorem we obtain $AC=\boxed{24}.$ ~samrocksnature

## Solution 2 (Stewart's Theorem)

Note that $\angle BDC = 180-x$, which means $\angle DCB = \angle DBC$ and $AD=DB=DC=13$. Now, Stewart's Theorem dictates $x^2 \cdot 13 = 7488$, yielding $AC=x=\boxed{24}$ ~Geometry285