Question

已知,如图①,在△ABC中,∠ABC=45°,H是两条高线AD和BE的交点.

Answer 1

1.∠EHA=∠BHD

∠HDC=∠HEA

∴180°-∠EHA-∠HEA=180°-∠BHD-∠HDC

即∠HBD=∠HAE

∠ABD=45°=∠BAD

AD=BD

∵∠HBD=∠HAE

∠BDH=∠ADC=90°

AD=BD

∴△BDH≌△ADC

∴BH=AC