Question

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∽

Answer 1

∠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