An isosceles triangle $ABC$ has $AB=AC$. Angle $A$ measures $20^\circ$. On $AC$, point $E$ is such that $AE=BC$. The task is to find the measure of angle $BEC$ without using trigonometry. How can one go about this?
Construct an isosceles triangle AEQ with ∠AQE = 20°. Since. AE = BC, the latter is equal to ΔABC. In particular, AQ = AB. Also,
∠BAQ = ∠EAQ - ∠EAB = 80° - 20° = 60°.
Which makes ΔABQ equilateral. In particular, BQ = EQ = AQ.