Question

在正方形ABCD中,CE垂直于∠CAB的平分线AE ,AE交BC于点F,求证:CE=1/2AF

Answer 1

证明：由题知,∠CAB＝45°,∠BAF＝∠CAF＝45°/2,AC＝AB×根号2

∴ AF＝AB/cos(45°/2),CE＝AC×sin(45°/2)

∴ CE/AF=[AC×sin(45°/2)]/[AB/cos(45°/2)]

＝[根号2×AB×sin(45°/2)]/[AB/cos(45°/2)]

＝根号2×sin(45°/2)×cos(45°/2)

＝1/2

∴ CE＝AF/2