# Thread: Find angle x interior point inside triangle

1. ## Find angle x interior point inside triangle

Isosceles triangle ABC (AB=AC) has interior point P such that angle PBC = 10 and PCB = 30. Angle BAC = 80. Find angle APB without using cosine/sine rule.

bump.

3. ## Re: Find angle x interior point inside triangle

You DO know that we don't do homework, right?

How far did you get with this problem?
Can you calculate angles ABP and ACP?

4. ## Re: Find angle x interior point inside triangle

I apologise, this isn't homework. I'm studying for an exam and this question came up and I needed some help solving it.

ABP and ACP are relatively simple to solve using the isosceles triangle, ABP = 40, ACP = 20.

I've tried drawing altitudes from P to AB and AC, constructing a cyclic quad, and all sorts of things such as constructing a perpendicular line from A to BC. None of which have enabled me to calculate the angle APB.

5. ## Re: Find angle x interior point inside triangle

Problem can be solved by coordinate geometry.Is this covered in your course.

6. ## Re: Find angle x interior point inside triangle

HINT:
show that BP = AB

Then you have isosceles triangle ABP with angle ABP = 40

7. ## Re: Find angle x interior point inside triangle

Denis,

I carefully drew a diagram a week ago and observed that triangle ABP is indeed isosceles. I've been trying to prove this, but can't do it. I give up; could you post your solution?

8. ## Re: Find angle x interior point inside triangle

BP meets AC at D. Bisector of angle BAC meets BD at O
O is the circumcenter of ABC, angle OCB=10
Let OC=OB=OA=R, AB=AC=b, AD=c, CD=b-c, and BD=d
We want to show BP=b
$\displaystyle \frac{R}{b}=\frac{d-R}{c}=\frac{c}{d}=\frac{R+(d-R)}{b+c}=\frac{d}{b+c}$
From this we get
$\displaystyle \frac{c}{d}=\frac{d}{b+c}$

$\displaystyle R=\frac{b c}{d}=\frac{b d}{b+c}$

$\displaystyle \text{OD}=d-R=\frac{d^2-b c}{d}=\frac{c^2}{d}$

CP bisects angle OCD, angle OCP = angle PCD = 20 = angle COP
Triangle OPC is isosceles and CP=OP

Similar triangles OCD, CPD

$\displaystyle \frac{\text{OC}}{\text{CP}}=\frac{\text{OD}}{\text {CD}}=\frac{R}{\text{OP}}=\frac{\text{OD}}{b-c}$

From this we get $\displaystyle \text{OP}=\frac{b(b-c)}{c}$

and

$\displaystyle \text{BP}=R+\text{OP}=\frac{b d}{b+c}+\frac{b(b-c)}{c}=\frac{b\left(c d+b^2-c^2\right)}{c(b+c)}=b$

For the last equality note that $\displaystyle b^2=c^2+d^2-c d$ in triangle ABD using the fact that angle ADB = 60 so we don't have to say 'law of cosines'

9. ## Re: Find angle x interior point inside triangle

Thank you so much for the solution.

I have one question, how did you determine/prove that circumcenter O lies on BD?

10. ## Re: Find angle x interior point inside triangle

Originally Posted by eskimogenius
Thank you so much for the solution.

I have one question, how did you determine/prove that circumcenter O lies on BD?
OAB iso since angle OBA = angle OAB = 40 so OA=OB

OB=OC since O is on the perpendicular bisector of side BC

OA=OB=OC