已知:在△ABC中,AB=AC,∠BAC和∠ACB的平分线AE、CD相交于D。∠ADC=130°。求∠BAC的度数

2024-12-25 16:17:28
推荐回答(1个)
回答1:

AB=AC
∠B=∠ACB
=1/2(180°-∠BAC)
=90°-1/2∠BAC
∠ACD=1/2∠ACB
=1/2(90°-1/2∠BAC)
=45°-1/4∠BAC
∠CAD+∠ACD+∠ADC=180°
1/2∠BAC+45°-1/4∠BAC+130°=180°
1/4∠BAC=5°
∠BAC=20°