Let $\displaystyle \bigtriangleup ABC$ be a right angled triangle such that $\displaystyle \angle A =90$°$\displaystyle , AB =AC$ and let $\displaystyle M$ 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.

(a) Find the ratio of the areas of the two triangles $\displaystyle \bigtriangleup ABH : \bigtriangleup AHM$.

(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 :

$\displaystyle AB : AM = 2:1$

$\displaystyle \bigtriangleup ABH : \bigtriangleup AHM$

$\displaystyle = 2^2 : 1^2 = 4 : 1$

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, $\displaystyle \bigtriangleup AHM = \bigtriangleup HCM$ --- (1)

on the other hand, $\displaystyle \frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC}$ --- (2)

From (1) and (2) $\displaystyle \frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{2\bigtriangleup AHM}{\bigtriangleup HPC}$ --- (3)

therefore from (3) and question (a)

$\displaystyle \frac{\bigtriangleup HPC}{\bigtriangleup HBP} = \frac{2\bigtriangleup AHM}{\bigtriangleup ABH} = \frac{(2)(1)}{4} = \frac{1}{2}$ --- (4)

$\displaystyle \frac{PC}{BP} = \frac{\bigtriangleup HPC}{\bigtriangleup HBP}$ 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.