Calculating Ratio of BP and PC in Right Triangle ΔABC

[songoku]

Homework Statement

let $$\Delta$$ABC be a right angled triangle such that angle A = 90^o, AB = AC and let 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. Find the ratio BP : PC

Homework Equations

The Attempt at a Solution

Please give me a clue to start

 

[dx]

Take the point P on the side BC so that AP is vertical to BM.

Do you mean perpendicular?

 

[g_edgar]

Please give me a clue to start

Draw a picture

 

[songoku]

Do you mean perpendicular?

yes it's perpendicular

i have drawn a picture

because CM = MA, is it correct if i assume that angle CMB = angle AMB = 22.5^o?

thx

 

[Дьявол]

If AB=AC then you got isosceles triangle and the angles are: 90,45,45 degrees. Also P divides BC on two equal parts (since the triangle is isosceles).

Using sine and the Pitagorean theorem you would easly come up with AH and AM. Since AM=AC/2, you will find AC.

Regards.

 

[songoku]

sorry but the question is BP : PC

if P divides BC on two equal parts, it means that BP : PC = 1 : 1
but the answer is 2 : 1 and i don't know how to get it

is angle CMB = angle AMB = 22.5 degree?

thx

 

[Leong]

1. A(0,0); B(a,0); C(0,a), M(0,a/2).
2. Slope of MB = -1/2
3. Slope of AP = 2
4. Equation of AP: y = 2x
5. Equation of CB : y = -x + a
6.Intersection of AP & CB is P(a/3, 2a/3).
7. BP: PC = 2:1.

 

[songoku]

wow, i never thought of using line equation and gradient
really nice...

thx a lot leong ^^