Question

已知：如圖所示，從Rt△ABC的兩直角邊AB，AC向外作正方形ABFG及ACDE，CF，BD分別交AB，AC于P，Q.求證：AP=AQ.

Answer 1

∵∠BAC+∠BAG=90°+90°=180°,
∴C,A,G三點(diǎn)共線.同理B,A,E三點(diǎn)共線.
∵AB∥GF,AC∥ED,∴=，=，
即AP=，AQ=.
又∵CA=ED=AE，GF=BA=AG，
∴CG=CA+AG=AE+BA=BE.
∴AP=AQ.

證明  ∵∠BAC+∠BAG=90°+90°=180°,
∴C,A,G三點(diǎn)共線.同理B,A,E三點(diǎn)共線.
∵AB∥GF,AC∥ED,∴=，=，
即AP=，AQ=.
又∵CA=ED=AE，GF=BA=AG，
∴CG=CA+AG=AE+BA=BE.
∴AP=AQ.