Finding an angle in a triangle where a median from B equals an altitude from A

Hi everyone,

This problem has me stumped as I cannot find an relation to use to solve the problem. The solution had me even more confused.

In triangle ABC, altitude AH and median BM intersect inside the triangle and are equal. If <ACB = 41 degrees, find the measure of <MBC.

The solution stated to draw perpendicular MK from M to line BC. In triangle AHC, MK = 1/2 AH, so MK = 1/2 BM, so <MBC = 30. I don't understand where the conclusion of MK = 1/2 AH and am also wondering if a simpler solution exists without having to draw line MK.

Re: Finding an angle in a triangle where a median from B equals an altitude from A

Quote:

Originally Posted by

**ShadowKnight8702** I don't understand where the conclusion of MK = 1/2 AH

That's because triangles MKC and AHC are similar with the factor of 1/2 since MC = (1/2)AC.

Re: Finding an angle in a triangle where a median from B equals an altitude from A

to further help consider the triangles AHC and MKC

AM = MC because BM is median Hence AC = 2 MC

angle AMC = angle MKC = 90 and angle C is common thus triangles AHC and MKC are similar

Thus their sides should be in proportion

Hence AH/MK = AC / MC

Now you can take it further