Question

如图,在△ABC中,AB=AC,点D、E分别在BC、AC的延长线上,AD=AE,∠CDE=30°求:∠BAD的度数

Answer 1

解：
在△ABC中,AB=AC
则有：∠ABC=∠ACB
且∠BAC=180°-2∠ACB
又在△ADE中,AD=AE
则：∠AED=∠ADE,
且∠CAD=180°-2∠AED
所以∠BAD=∠BAC+∠CAD
=180°-2∠ACB+180°-2∠AED
=360°-2(∠ACB+∠AED)
又∠ACB=180°-∠BCE,∠BCE=∠AED+∠CDE=∠AED+30°
所以：∠ACB=180°-(∠AED+30°)=150°-∠AED
即：∠ACB+∠AED=150°
所以：∠BAD＝360°-2(∠ACB+∠AED)
＝360°－300°
＝60°