試題詳細解析教學---**Problem Description:** 頂角為20度的等腰三角形△ABC，做AB邊上之E點滿足AE=BC。 An isosceles triangle △ABC with vertex angle 20°. Point E is made on side AB such that AE = BC. Given: △ABC is an isosceles triangle. ∠BAC = 20° Point E is on side AB. AE = BC **Other Relevant Text:** ∠ABC = 80° (This is consistent with △ABC being isosceles with ∠BAC = 20° and AB=AC, as (180 - 20)/2 = 80) AE = BC (Written as AE = BC) **Question:** 試求∠BEC = ? Find ∠BEC = ? **Geometric Figure Description:** * Type: Geometric figure (Triangle). * Main Elements: * Triangle ABC with vertices labeled A, B, C. A is at the top, B is bottom left, C is bottom right. * Angle at vertex A (∠BAC) is marked with an arc and labeled 20°. * Angle at vertex B (∠ABC) is marked with an arc and labeled 80°. * Side AB and AC are marked with double vertical hash marks, indicating AB = AC. * Side BC is marked with double horizontal hash marks. * Point E is located on side AB, between A and B. * Segment AE is marked with double horizontal hash marks, indicating AE = BC. * Segment EC is drawn, forming triangle BEC. * The triangle ABC area is filled with orange color. * The arc for ∠ABC is filled with gray color. * The arc for ∠BAC is partially visible near A.