∠BDC=∠DBM+∠BAD+∠DAC+∠ACD
=1/2∠B+∠BAC+1/2∠C
=1/2(∠B+∠BAC+C)+1/2∠BAC
=90+1/2∠BAC
∠BMD=90+1/2∠MAC
所以∠BDC=∠BMD
又∠DBC=∠MBD
所以 :△MBD∽△DBC
同理：:△MBD∽△DNDC
相似三角形证明
△ABC中,三内角的平分线交于点D,过D作AD的垂线分别交AB、AC于点M、N,求证:△MBD∽△DBC∽NDC
∠BDC=∠DBM+∠BAD+∠DAC+∠ACD
=1/2∠B+∠BAC+1/2∠C
=1/2(∠B+∠BAC+C)+1/2∠BAC
=90+1/2∠BAC
∠BMD=90+1/2∠MAC
所以∠BDC=∠BMD
又∠DBC=∠MBD
所以 :△MBD∽△DBC
同理：:△MBD∽△DNDC