Question

等腰△ABC中,顶角A=100°,作∠B的平分线交AC于点E,求证：BC=AE+EB

Answer 1

AB=AC 且∠A=100°则∠B=∠C=40°如图所似;∠AEB=60°∠BEC=120°

AE=AC=1/2AB;(1)

BC^2=BE^2+CE^2-2BE*CE*cos120°

AB^2=AE^2+BE^2-2AE*BE*cos60°;

代入(1)式

可得BC=AE+EB;