三角形abc中,角acb=90度,ac=bc,ae是bc边上的中线,过点c做cf垂直ae,垂足为f,过点b作bd垂直bc,交cf的延长线于点d.是说明ae=cd

Question

三角形abc中,角acb=90度,ac=bc,ae是bc边上的中线,过点c做cf垂直ae,垂足为f,过点b作bd垂直bc,
交cf的延长线于点d.是说明ae=cd

Answer 1

证明：如图,∵CF⊥AE
                     ∴∠1+∠3=90°
                   又∵∠ACB=90°
                    ∴∠2+∠3=90°
                     ∴∠1=∠2
                  ∵BD⊥BC
                ∴ ∠CBD=90°
            在Rt△ACE和Rt△CBD中
          ∵ ∠ACE=∠CBD=90°
                            AC=BC
                         ∠1=∠2
                    ∴Rt△ACE≌Rt△CBD （角、边、角）
             ∴AE=CD