已知等腰直角三角形ABC,P是斜边BC上一点,求证2AP²=BP²+PC²过点A作AD⊥BC于点D∵AB⊥AC,AB=AC∴AD=BD=CD∴BP²+PC² =(BD-PD)²+(BD+PD)² =BD²-2BD·PD+PD²+BD²+2BD...