Question

In $\triangle ABC$, we have $AB = AC = 13$ and $BC = 10$. Let $M$ be the midpoint of $\overline{AB}$ and $N$ be on $\overline{BC}$ such that $\overline{AN}$ is an altitude of $\triangle ABC$. If $\overline{AN}$ and $\overline{CM}$ intersect at $X$, then what is $AX$?

Answer 1

since AB=AC, ABC is isosceles, and AN, besides being an altitude, is another median, like CM.

The medians intersect 2/3 of the way from each vertex to the opposite side.

The altitude AN forms two 5-12-13 right triangles, so AN=12, and AX=8.