Originally Posted by

**Unknown008** Okay, to come to my solution about it, I used variables instead of giving specific values.

I let HM = x, this means that AH = 2x (due to similar triangles and I know that 2AM = AB) and using Pythagoras, I get AM = $\displaystyle x\sqrt{5}$

From this, we get AB = $\displaystyle 2x\sqrt{5}$ and HB = 4x.

From there, I can get angles BAH and HAP which are respectively $\displaystyle \sin^{-1}\left(\frac{4x}{2x\sqrt{5}}\right)$ and $\displaystyle \sin^{-1}\left(\frac{x}{x\sqrt{5}}\right)$ respectively.

In degrees those give 63.43 and 26.57 degrees (I kept all the other numbers to retain the accuracy as much as possible).

And then I get angles ABP = 45, ACP = 45, APB = 180-(45+63.43) = 71.57, APC = 180-(45+26.57) = 108.43 degrees.

Since the sides don't matter now, I put AB = 2 cm and AH = 1 cm. And from those and using the angles I just found, using the sine rule, I get side BP = 2.12 cm and PC = 0.943 cm. The ratio then turns out to be 9:4