- #1

Vahn

- 2

- 0

## Homework Statement

Let [itex]\bigtriangleup[/itex] ABC be a right angled triangle such that [itex]\angle[/itex] 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 [itex]\frac{AB\sqrt{2}}{PC} - 1[/itex], 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 [itex]\frac{BP}{PC} = \frac{BC'}{AC'}[/itex] and then, using the intercept theorem, [itex]\frac{BP}{BC'} = \frac{PC}{AC'} = \frac{BC}{AB} = \sqrt{2}[/itex]. I can't figure out what to do with that information, so it's another dead end.