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.
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.
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
Similar triangles OAD, ABD gives
$\displaystyle \frac{R}{b}=\frac{d-R}{c}=\frac{c}{d}=\frac{R+(d-R)}{b+c}=\frac{d}{b+c}$
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'