# solving an isosceles triangle

My kid has this problem assigned and hasn't yet learned the laws of sines and of cosines so can't be expected to use them. (The use of trig functions in a right triangle has been covered, however.)

ABC is a triangle with AB=AC. E is the midpoint of AC; D is the midpoint of BC. BE and AD meet at point N. BE=12; AD=15. Find the measure of angle BND.

I thought of using the trapezoid AEDB, the isosceles triangle AED, and/or the similar triangles CED~CAB, but don't see how any of those helps. I'm probably missing something obvious, and would appreciate your help.

• Sorry for the lack of a drawing: I'm on a phone. – msh210 Oct 24 '17 at 19:34
• Are you allowed to use trigonometric functions? – Mr. T Oct 24 '17 at 19:37
• @Mr.T, I've edited to clarify. Thanks. – msh210 Oct 24 '17 at 19:39

Hint: medians in the triangle intersect each other in the ratio 2:1. Also, $AD$ is a height.

• Ah, that first fact is what I was missing. And probably my kid has learned it. Thank you!! – msh210 Oct 24 '17 at 19:41
• Yes, this fact should be well known, it follows from the triangle midpoint theorem which I was explaining to my son a week ago :) – Vasya Oct 24 '17 at 19:43

The three medians of a triangle trisect each other at their point of concurrency. In this case, the given lengths $AD$ and $BE$ are medians meeting at N:

$cos^{-1}(\frac{5}{8})$ is the measure of angle $BND$.