AH⊥BC交BC于E 连接BH
∠HBC=∠HAC=90°-∠C
∠HDB=∠DHE+90°
∠AHN=∠EHM=∠DHE+90°
∠HDB=∠AHN
△HDB∽△AHN
HN/BD=AH/HE
同理可证
△HDC∽△AHM
HM/CD=AH/HE
HM/CD=HN/BD
BD=CD
HM=HN
AH⊥BC交BC于E 连接BH
∠HBC=∠HAC=90°-∠C
∠HDB=∠DHE+90°
∠AHN=∠EHM=∠DHE+90°
∠HDB=∠AHN
△HDB∽△AHN
HN/BD=AH/HE
同理可证
△HDC∽△AHM
HM/CD=AH/HE
HM/CD=HN/BD
BD=CD
HM=HN