SOLUTION: Hi! I'm stuck trying to solve this problem: https://i.postimg.cc/cJgtCnbg/Image-148.jpg The answers are: ABH : AHM = 4 BP : PC = 2 This is a sketch that comes with t

Algebra ->  Triangles -> SOLUTION: Hi! I'm stuck trying to solve this problem: https://i.postimg.cc/cJgtCnbg/Image-148.jpg The answers are: ABH : AHM = 4 BP : PC = 2 This is a sketch that comes with t      Log On


   

Question 1136012: Hi!
I'm stuck trying to solve this problem:
https://i.postimg.cc/cJgtCnbg/Image-148.jpg
The answers are:
ABH : AHM = 4
BP : PC = 2
This is a sketch that comes with the question:
https://i.postimg.cc/V6sy4qP5/Image-149.jpg
I figured out the first part by using the properties of right triangles on ABM.
I don't know how to proceed in regards to the second part of the question, though.
Any help would be welcome. Thanks in advance!
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

its a known property that such medians divide the triangle sides in the ratio of 2%3A1+
so,
sides BP+%3A+PC+=+2%3A1

since sides are split in the ratio of 2%3A1, and since right triangle is isosceles, their area will be in the ratio of 4%3A1

let the areas of two triangles ABH and AHM be area 1+ and area 2+

since area 1+=+%281%2F2+%29%2Ab%2Ah+=+bh%2F2+
area 2+=+%281%2F2%29+%2A%282b%29+%2A+%282h%29+=+2bh+
area_+2+%3A+area_+1+=+2bh+%2F+%28bh%2F2%29+=+4+bh%2Fbh+=+4+%2F+1+=+4%3A1