如图所示,三角形ABC中,角A=108度,AB=AC,BD是角平分线.求证:BC=AB CD.

问题描述:

如图所示,三角形ABC中,角A=108度,AB=AC,BD是角平分线.求证:BC=AB CD.

是证明 BC =AB+CD证明在BC上取点E,使CE=CD,连接DE∵ ∠A=108°,AB=AC∴ ∠C=36°∵ CE=CD∴ ∠DEC=72°∴ ∠DEB=180-72=108°∵ ∠A=108°∴ ∠DEB=∠A∵ BD是∠B的平分线∴ ∠EBD=∠ABD∵ ∠DEB=∠A,BD=BD∴ △BE...