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
From this we get
CP bisects angle OCD, angle OCP = angle PCD = 20 = angle COP
Triangle OPC is isosceles and CP=OP
Similar triangles OCD, CPD
From this we get
and
For the last equality note that in triangle ABD using the fact that angle ADB = 60 so we don't have to say 'law of cosines'