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

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 $AC=\boxed{24}.$

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

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2021 JMPSC Invitationals Problems/Problem 9

Contents

Problem

Solution

Solution 2 (Stewart's Theorem)

See also