# Triangle question

1. Jun 29, 2009

### songoku

1. The problem statement, all variables and given/known data
let $$\Delta$$ABC be a right angled triangle such that angle A = 90o, 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

2. Relevant equations

3. The attempt at a solution
Please give me a clue to start

2. Jun 29, 2009

### dx

Do you mean perpendicular?

3. Jun 29, 2009

### g_edgar

Draw a picture

4. Jun 30, 2009

### songoku

yes it's perpendicular

i have drawn a picture

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

thx

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5. Jun 30, 2009

### 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.

Last edited: Jun 30, 2009
6. Jul 1, 2009

### 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

7. Jul 1, 2009

### 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.

8. Jul 2, 2009

### songoku

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

thx a lot leong ^^