问题补充:
已知如图在三角形abc中,角bac等于90度,ab=ac,ad⊥bc于d,e为ac上一点,be交ad于h,af⊥be于g求证:dh=dfhttps://d./zhidao/wh%3D600%2C800/sign=42e8b736d11373f0f56a6799943f67c3/6d81800a19d8bc3e229e2987828ba61ea9d345ec.jpg
答案:
∵ 角bac等于90度,ab=ac,ad⊥bc
∴ AD三线合一 AD=BD
∵ ad⊥bc af⊥be
∠DAF+∠AFD =90 ∠HBD+∠AFD=90
∴ ∠DAF=∠HBD
∵ ∠ADF=∠BDH=90
∴ △ADF ≌△BDH
∴ DH=DF