+0

# triangles

+1
1110
1
+75

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 $AX$?

Jul 20, 2017

#1
+94550
+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

Jul 20, 2017
edited by CPhill  Jul 21, 2017