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 $AX$?

eileenthecoolbean Jul 20, 2017

#1**+2 **

Orient the triangle according to the following pic :

ABC is isosceles....altitude AN forms leg 12 of a 5-12-13 right triangle

The midpoint of AB = M = (2.5, 6)

From M, draw a perpendicular to BC.....call this DM = 6

And because DM is parallel to the altitude...triangles CDM and CNX will be similar

So

DM / CD = NX / CN

And CD = BC - BD = 10 - 2.5 = 7.5

And CN = 1/2 of BC = 5

So

6 / 7.5 = NX / 5

(5 * 6 ) / 7.5 = NX

30 / 7.5 = NX

4 = NX

But AN - NX = AX

So 12 - 4 = AX = 8

CPhill Jul 20, 2017