Question

Let ABC be a right angled triangle such that /_ A = 90º , AB = AC and let M be the mid point of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM.

1 - Find the ratio of the areas of the two triangles ABH : AHM .
2- Find the ratio BP : PC

Answer 1

In the first part, I could find that AM = AB/2 , but how can I apply it on the question?

Answer 2

@Styxer
Need help?

Answer 3

If AP is vertical to BM means AP perpendicular to BM, then
AP and BM are 2 "medianes then H is the center of gravity and stay at 2/3 of each "medianes" from the vertex. So, BH=2*HM and ABH/AHM=2.
BP/PC=1 (P is the middle of BC

Answer 4

@3mar yes please

@caylus Both of your answers aren't correct, according to the answer sheet :c

Answer 5

That is with my pleasure!

Did you try to draw that triangle?

Answer 6

Yes, hold on, I'll upload the photo

Answer 7

That would be better!

Answer 8

They are out of scale, btw

Answer 9

That does not matter!..

What matters is that we do the math correctly!

Answer 10

I can't see where the AB = AM/2 can be used, because we would use (base*height)/2 to find the areas... or (AB*AH*sin 90)/2 ... Both don't use AB or AM

Answer 11

sorry it's (AH*HB*sin 90)/2

Answer 12

I think that is what you got..