Question

如图,在△ABC中,AD平分∠BAC,P为线段AD上的一个动点,PE⊥AD交直线BC于点E,当P点在线段AD上运动时,猜

Answer 1

∠E=90°-∠ADC=90°-(∠B+∠BAD)=90°-(∠B+∠BAC/2)

=90°-(∠B+∠BAC/2)=90°-[∠B+(180°-∠B-∠ACB)/2]=90°-(∠B+90°-∠B/2-∠ACB/2)

=(∠ACB-∠B)/2