解(证明):在$\triangle ABC$和$\triangle ADC$中,$\begin{cases}AB = AD\\CB = CD\\AC = AC\end{cases}$所以$\triangle ABC\cong\triangle ADC$($SSS$)。则$\angle BAC=\angle DAC$。设$AC$与$BD$相交于点$O$。在$\triangle ABO$和$\triangle ADO$中,$\begin{cases}AB = AD\\\angle BAO=\angle DAO\\AO = AO\end{cases}$所以$\triangle ABO\cong\triangle ADO$($SAS$)。则$\angle AOB=\angle AOD$。又因为$\angle AOB+\angle AOD = 180^{\circ}$,所以$\angle AOB=\angle AOD = 90^{\circ}$,即$AC\perp BD$。