Question

已知三角形ABC中角BAC=90度,AB=AC,CD垂直于角ABC的平分线BD于D,BD交AC于E,求证BE=2CD

Answer 1

证明：延长BA、CD交于点F.

∠ABE+∠AEB=∠ECD+∠CED=90°∠AEB=∠CED,∴∠ABE=∠ECD.

又有∠BAE=∠CDE,AB=AC ∴△ACF≌△ABE,BE=CF

BD⊥AC且平分∠ABC,所以△FBC是等腰三角形,BD也是底边上中线.CF=BE=2CD