Let be a right angled triangle such that ° and let 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.
(a) Find the ratio of the areas of the two triangles .
(b) Find the ratio BP : PC.
Now this is my triangle, to place P, I took "AP is vertical to BM" as "AP and BM are perpendicular".
Now the answer to (a) is :
It is true that AB is twice large than AM, but why are they rising them to the 2th power?
To solve (b) I tried measuring PB and CP, and the comparing with CB, but I gave up soon, as I was getting something near 2.3 for CP. so I looked at the answer (2:1), and noticed the should be:
CB = 7
CP = 2.333333333
PB = 4.666666666
Though method used to calculate the ratio is very strange, I don't understand it:
M is in the middle of AC, --- (1)
on the other hand, --- (2)
From (1) and (2) --- (3)
therefore from (3) and question (a)
because of (4)
The answer is BP : CP = 2 : 1
Can somebody explain me how to use this to get the ratio?
Thanks in advance.