已知:平行四边形ABCD中,AB=3,BC=5.
求:AC²+BD²的值.
解:作AE垂直BC于E,DF垂直BC的延长线于F.则AE平行DF.
∵∠ABE=∠DCF;AB=DC;∠AEB=∠DFC=90°.
∴⊿ABE≌⊿DCF(AAS),BE=CF.
设BE=CF=X,AE=DF=Y,则:
AC²+BD²=(CE²+AE²)+(BF²+DF²)=(5-X)²+Y²+(5+X)²+Y²=50+2(X²+Y²)=50+2*AB²=68.