如图,在三角形ABC中,AB=AC,∠BAC=90°,在AB上取点P.在CA的延长线上取点Q,使AP=AQ,边CP与BQ交与点S,

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如图,在三角形ABC中,AB=AC,∠BAC=90°,在AB上取点P.在CA的延长线上取点Q,使AP=AQ,边CP与BQ交与点S,
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如图,在三角形ABC中,AB=AC,∠BAC=90°,在AB上取点P.在CA的延长线上取点Q,使AP=AQ,边CP与BQ交与点S,
如图,在三角形ABC中,AB=AC,∠BAC=90°,在AB上取点P.在CA的延长线上取点Q,使AP=AQ,边CP与BQ交与点S,

如图,在三角形ABC中,AB=AC,∠BAC=90°,在AB上取点P.在CA的延长线上取点Q,使AP=AQ,边CP与BQ交与点S,
证明:
∵∠BAC=90
∴∠BAQ=180-∠BAC=90
∴∠BAQ=∠BAC
∵AB=AC,AP=AQ
∴△CAP≌△BAQ (SAS)