1、在ΔABC中,AD⊥BC于点D∠B=2∠C.试说明AB+BD=CD.
证明：在DC上取点E,使DE=BD,连接AE,
∵AD⊥BC,∴AD垂直平分BE,∴AB=AE,∠B=∠AEB,
∵∠B=2∠C,∠AEB=∠C+∠EAC,
∴2∠C=∠C+∠EAC,∴∠C=∠EAC,∴AE=CD,
∵CD=CE+DE=AE+BD,
∴CD=AB+BD.
2、在ΔABC中,AD⊥BC于点D,AB+BD=CD,.试说明∠B=2∠C.
证明：在DC上取点E,使DE=BD,连接AE,
∵AD⊥BC,∴AD垂直平分BE,∴AB=AE,∠B=∠AEB,
∵CD=AB+BD,CD=DE+CE,∴CE=AB=AE,
∴∠EAC=∠C,
∵∠AEB=∠EAC+∠C=2∠C,
∴∠B=2∠C.