Question

如图,在△ABC中,∠ACB=90°,CD⊥AB于点D,∠CAB的平分线AE分别交于BC和CD于点E、F.请说明CE=CF

急急急 必须今天有答复

Answer 1

∠FCE=90-∠B=∠A
∠FEC=90-∠A/2
∠EFC=180-∠FCE-∠FEC=180-∠A-(90-∠A/2)=90-∠A/2
=>∠EFC=∠FEC
=>CE=CF

Answer 2

图跑哪去了
八嘎