In triangle ABC BM median and BH height it is known that AC = 17 and BC = BM find AH.

Since BM is the median, BM divides the AC side in half with the point M: AM = MC. Let’s find the length of the MC:
MC = AC / 2;
MC = 17/2 = 8.5 (conventional units).
Consider the MBC triangle. MBC is an isosceles triangle, since BM = BC (by condition), then the height BH in an isosceles triangle MBC is both a bisector and a median (properties of an isosceles triangle). Then, BH, as the median, by the point H divides the side of the triangle MVC MC in half: MH = HC. Let’s find the length of MH:
MH = MC / 2;
MH = 8.5 / 2 = 4.25 (conventional units).
The segment AH to be found consists of two segments AM and MH. Let’s find the length АН:
AH = AM + MH;
AH = 8.5 + 4.25 = 12.75 (conventional units).
Answer: An = 12.75 conventional units.