Triangle's elementary geometry problem

[Vahn]

Homework Statement

Let \bigtriangleup ABC be a right angled triangle such that \angle A = 90°, AB = AC and let M be the midpoint 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. Find the ratio BP:PC.

Homework Equations

Possibly the intercept theorem and others related to the congruency of line segments and triangles similarity.

The Attempt at a Solution

I first tried to attack the problem using vectors, but my limited knowledge of their rules meant I quickly found a dead-end. I then tried expressing BP as BC-PC, and since BC is the hypotenuse, the ratio is \frac{AB\sqrt{2}}{PC} - 1, but could not figure how to express PC as a function of AB.

I then spent some days trying to fiddle with proportions to no avail. My last try involved connecting the points C and H, and then extending the resulting line segment until it crossed AB, and called that point C'. By Ceva's theorem, I found out that \frac{BP}{PC} = \frac{BC'}{AC'} and then, using the intercept theorem, \frac{BP}{BC'} = \frac{PC}{AC'} = \frac{BC}{AB} = \sqrt{2}. I can't figure out what to do with that information, so it's another dead end.

 

[HallsofIvy]

What I would do is set up a coordinate system. Take A to be at (0, 0), B at (0, 1), and C at (1, 0). The M is at (1/2, 0) and it is easy to find the equation of line BM and then line AP. Use that to find point P.

 

[Vahn]

I see. So, assuming C = (0,0), A = (1, 0), B = (1, 1), I have:

BM: y_{BM} = 2 x_{BM} - 1
AP: y_{AP} = \frac{- x_{AP} + 1}{2}
BC: y_{BC} = x_{BC}

So P = (\frac{1}{3}, \frac{1}{3}) and BP:PC=2:1, which is the correct answer :).

However, the 'official' resolution of this problem is the following:

1 - \bigtriangleup AHM = \bigtriangleup HCM

2 - \frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC}

3 - \frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{2 \bigtriangleup AHM}{\bigtriangleup HPC} = \frac{1}{2}

4 - \frac{BP}{PC} = \frac{\bigtriangleup HBP}{\bigtriangleup HPC} = 2

I'm completely lost.

 

[LCKurtz]

You can do it with just geometry. At C draw a line parallel to AP meeting BA extended at W. Then triangle AWC is similar to AMB. This gives AW/AM = AC/AB = 2AM/AB or

AW = 2 AM²/AB

Now BP/PC = BA/AW. Substitute AW from the above equation into this giving

BP/PC =(1/2)(BA/AM)²= (1/2)*4 = 2