如图,已知在三角形ABC中,角C=90°,D为AC上的一点,AB的平方-BD的平方与AC的平方-DC的平方有什么关系?

(BC)^2+(DC)^2=(BD)^2.(1)
(BC)^2+(AC)^2=(AB)^2.(2)
(AC)=(AD)+(CD).(3)
(AB)^2-(BD)^2=[(BC)^2+(AC)^2]-[(BC)^2+(DC)^2]
={(BC)^2+[(AD)+(DC)]^2}-[(BC)^2+(DC)^2]
={(BC)^2+[(AD)^2+2(AD)(DC)+(DC)^2]}-(BC)^2-(DC)^2
=(BC)^2+(AD)^2+2(AD)(DC)+(DC)^2-(BC)^2-(DC)^2
=(AD)^2+2(AD)(DC)
(AC)^2-(DC)^2=[(AD)+(DC)]^2-(DC)^2
=[(AD)^2+2(AD)(DC)+(DC)^2]-(DC)^2
=(AD)^2+2(AD)(DC)
因此[(AB)^2-(BD)^2]=[(AC)^2-(DC)^2]