Question

ABC是等腰三角形,分别以它的两腰为边作等边三角形△ADB和△ACE,已知∠DAE=∠DBC分别求出ABC三个内角的度数

Answer 1

设∠A为x° 则∠DAE=60°+X°+60° ∠DBC =60°+∠ABC

得120°+X°=60°+∠ABC 又∠ABC=∠ACB=（180°-∠A）／2=90°-½X°

则120°+X°=60°+90°-½X° 解得X=20 ∠A=20° ∠ACB=∠ABC=80°