Question

1,.如图1,已知△ABC是等边三角形,BD是高,延长BC到点E,使CE=CD,过点D作DF⊥BE于点F.求证：F为线段

Answer 1

△ABC是等边三角形,BD是高 ==>BD为三线合一,

D为AC中点,∠DBC=30, ∠FDC=180-60-90=30,

CD=CE,∠CDE=∠CED,∠DCF=∠CDE+∠CED=60,

SO THAT ,∠CDE=∠CED=30=∠DBC,

DF为高,

证得 DB=DE,

从而,F必为BE中点