已知矩形ABCD的对角线AC和BD交于点O,在BC上取BE=BO,若∠BOE=75°,求∠CAE

题目

已知矩形ABCD的对角线AC和BD交于点O,在BC上取BE=BO,若∠BOE=75°,求∠CAE

答案

∠BOE=75°,BE=BO =>∠DBC=30°
=>∠ABD=∠BAC=60°
所以,三角形ABO是个等边三角形
=>AB=OB=BE =>三角形ABE是个等腰直角三角形
=>∠BAE=45°
=>∠CAE=60°-45°=15°