△ABC and CA = CB, ∠C = 20°.

Also, BD = DC and BE = BA.

Show that ∠BDE = 30°.

Hint: With center B and radius BD, construct an arc of a circle meeting BA (extended) at X and BC at Y. Draw the segments DX, DY; ponder.

Here’s the sketch, with the hint drawn. BX = BY (radii of the arc).

We want D3 = 30°. We will find D2, D4, and D5, then subtract.

- △BDC is isosceles → B2 = 20° and B1 = 60°. A1 = 80°.
- △BDY is isosceles → Y1 = D3 + D4 = 80° → Y2 = 100° → D5 = 60°.
- △XBD is isosceles with B1 = 60° → △XBD is equilateral → ∠X = 60° and XD = BD = DC.
- △XAD ≅ △DYC by ASA (∠X = D5, XD = DC, D1 = ∠C (because A2 = 100°) = 20°.
- XA = DY, CPCTC.
- And XA = EY (BX – BA = BY – BE) → DY = EY.
- △DEY is isosceles, Y2 = 100° (step 2) → E2 = D4 = 50°.
- D3 = 180° – D2 – D4 – D5 = 180° – 40° – 50° – 60° = 30°, q.e.d.