Triangle ABC is isosceles with base AC. Points P and Q are respectively in CB and AB and such that AC = AP = PQ = QB. What is the degree of angle B?
I have the solution, but I don't understand it.
Represent the magnitude of angle B by m. Then, in order, we obtain angle QPB = m, angle AQP = 2m (??? from now on), angle QAP = 2m, angle QPA = 180 - 4m, angle APC = 3m, angle ACP = 3m. Since angle BCA = angle BAC = 3m, the sum of the angles in ABC is m + 3m + 3m = 7m = 180.