Question

△ABC中,AB=AC,AD⊥BC于点D,DE⊥AC于点E,M是DE的中点,BE与AM交于点N,求证:角EBC=角DAM

Answer 1

证明:∵AB=AC;AD⊥BC.

∴CD=BC/2.

∵∠CDE=∠DAE(均为角ADE的余角);∠DEC=∠AED=90º.

∴⊿DEC∽⊿AED,CD/DA=CE/DE;

又CD=BC/2;DM=DE/2.

则(BC/2)/DA=CE/(2DM),BC/DA=CE/DM,BC/CE=DA/DM;

∵∠C=∠ADM(均为∠CDE的余角).

∴⊿BCE∽⊿ADM.(两边对应成比例且夹角相等的两个三角形相似)

故:∠EBC=∠DAM.