The diagram shows a triangle ABC with side BA extended to a point E. The bisector of ∠ABC meets the bisector of angle ∠EAC at D.
Let ∠BCA= p◦ and ∠BDA = q◦.
Prove that p = 2q.
Let a◦ = ∠EAD = ∠DAC and b◦ = ∠ ABD = ∠DBC. Points E, A, B are collinear, so ∠CAB = (180 − 2a)◦.
Adding up the angles in triangle ADB:
a + (180 − 2a) + q + b = 180
∴ q = a − b
Similarly, considering the angles in triangle ACB:
(180 − 2a)+ p + 2b = 180
∴p = 2a − 2b = 2q.
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