问题补充:
如图,△ABC中,AD是∠BAC的平分线,直线EF⊥AD,分别与AB、AC及BC的延长线交于点E、F、K,求证:∠K=12
答案:
证明:∵AD平分∠BAC,
∴∠BAD=∠DAC=12
======以下答案可供参考======
供参考答案1:
∠M=90°-∠ADM
=90°-(∠B+∠1)、
=90°-(∠B+1/2∠BAC)
=90°-【∠B-1/2(∠180°-∠B-∠ACB)}
=90°-(90°+1/2∠B-1/2∠ACB)
=1/2(∠ACB-∠B)
时间:2023-05-31 00:10:43
如图,△ABC中,AD是∠BAC的平分线,直线EF⊥AD,分别与AB、AC及BC的延长线交于点E、F、K,求证:∠K=12
证明:∵AD平分∠BAC,
∴∠BAD=∠DAC=12
======以下答案可供参考======
供参考答案1:
∠M=90°-∠ADM
=90°-(∠B+∠1)、
=90°-(∠B+1/2∠BAC)
=90°-【∠B-1/2(∠180°-∠B-∠ACB)}
=90°-(90°+1/2∠B-1/2∠ACB)
=1/2(∠ACB-∠B)