Question

已知:如图,在△ABC中,AB=AC,D为△ABC外一点,∠ABD=60°,AB=BD+DC,求证:∠ACD=60°

Answer 1

证明：如图,延长BD至E,使得DE=DC,连接AE.

因为AB=BD+DC,又DE=DC,所以AB=BD+DE+BE,又∠ABD=60°,所以△ABE是等边三角形,故AE=AB=AC,∠ABE=∠AEB=60°.在△ACD和△AED中,AC=AE,CD=BD,AD=AD,所以△ACD≌△AED,所以∠AEB=∠ACD=60°.