如图,点BAG上,点CAH上,BP平分∠GBC,CP平分∠HCB,PD⊥AG于DPE⊥AH于E求证:PD=PE.如图,点BAG上,点CAH上,BP平分∠GBC,CP平分∠HCB,PD⊥AG于DPE⊥AH于E求证：PD=PE.

如图,点BAG上,点CAH上,BP平分∠GBC,CP平分∠HCB,PD⊥AG于DPE⊥AH于E求证:PD=PE.如图,点BAG上,点CAH上,BP平分∠GBC,CP平分∠HCB,PD⊥AG于DPE⊥AH于E求证：PD=PE.
如图,点BAG上,点CAH上,BP平分∠GBC,CP平分∠HCB,PD⊥AG于DPE⊥AH于E求证:PD=PE.
如图,点BAG上,点CAH上,BP平分∠GBC,CP平分∠HCB,PD⊥AG于DPE⊥AH于E求证：PD=PE.

如图,点BAG上,点CAH上,BP平分∠GBC,CP平分∠HCB,PD⊥AG于DPE⊥AH于E求证:PD=PE.如图,点BAG上,点CAH上,BP平分∠GBC,CP平分∠HCB,PD⊥AG于DPE⊥AH于E求证：PD=PE.
作PQ⊥BC于Q,
∠PBD=∠PBQ,∠PDB=∠PQB=90°,PB=PB
易得RT△PDB≌RT△PQB [AAS]
PD=PQ
同理
RT△PEC≌RT△PQC [AAS]
PE=PQ
∴PD=PE