Question 1136025: Hi! There is this problem:
https://i.postimg.cc/cJgtCnbg/Image-148.jpg
Here's a sketch that was provided:
https://i.postimg.cc/V6sy4qP5/Image-149.jpg
Now, my request is not to solve the problem outright. The answer key outlines a solution, where it is asserted that:
https://i.postimg.cc/26H7t8wF/Image-150.jpg
What I want to know is, where does that formula come from? I tried to look it up, but to no avail. I know that area ratios of similar triangles are connected to the scale factor, but what theorem brings together the areas of these four triangles?
I'm sorry for the strange request. Thanks in advance!
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
I can see the justification for the proportion they show relating areas of triangles: ABH/HBP = AHC/HPC.
I don't see yet how the use of that leads to the answers to the questions that are asked.
Let J and K be on BC so that AJ and HK are perpendicular to BC.
Then using BC (or parts of BC) for the bases, AJ is an altitude of both BAP and PAC, and HK is an altitude of both BHP and PHC.
Then, since BAP and BHP both have base BP, and since PACand PHC have base PC, the ratio of the areas of BAP to BHP is the same as the ratio of the areas of PAC and PHC.
It then follows that the ratio of the areas of ABH and HBP is the same as the ratio of the areas of AHC and HPC.
While I don't see the path from there to the answers to the questions that are asked, the problem is easily solved using coordinate geometry.
Here is a sketch, with carefully chosen coordinates for the vertices of the original triangle ABC.
M is the midpoint of AC, so M = (45,15).
Then the equation of M is y = (1/3)x.
Line AP is perpendicular to BM, so its slope is -3; since it passes through A(30,30), the equation of AP is y = -3x+120.
Point P is the x-intercept of AP, so P is (40,0).
Point H is the intersection of BM and AP; solving the pair of equations for those lines finds H is (36,12):
Now the questions are easily answered.
(1) ratio of the areas of triangles ABH and AHM
Both triangles have altitude AH; and BH is 4/5 of BM, so the ratio of BH to HM -- and therefore the ratio of the areas of ABH and AHM -- is 4:1.
(2) ratio BP:PC
Simple -- 40:20 = 2:1
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