+77 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 $AX$?
+129933
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
