$ABCD$ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see figure). Prove that:
$(i)$ $\Delta ABD \cong \Delta BAC$
$(ii)$ $BD = AC$
$(iii)$ $\angle ABD = \angle BAC$
$(i)$ $\Delta ABD \cong \Delta BAC$
$(ii)$ $BD = AC$
$(iii)$ $\angle ABD = \angle BAC$
(N/A) $(i)$ In quadrilateral $ABCD$,we have $AD = BC$ and $\angle DAB = \angle CBA$.
Consider $\Delta ABD$ and $\Delta BAC$:
$AD = BC$ (Given)
$AB = BA$ (Common side)
$\angle DAB = \angle CBA$ (Given)
Therefore,by $SAS$ (Side-Angle-Side) congruence criterion,we have $\Delta ABD \cong \Delta BAC$.
$(ii)$ Since $\Delta ABD \cong \Delta BAC$,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = AC$.
$(iii)$ Since $\Delta ABD \cong \Delta BAC$,their corresponding parts are equal $(CPCT)$.
Therefore,$\angle ABD = \angle BAC$.
Consider $\Delta ABD$ and $\Delta BAC$:
$AD = BC$ (Given)
$AB = BA$ (Common side)
$\angle DAB = \angle CBA$ (Given)
Therefore,by $SAS$ (Side-Angle-Side) congruence criterion,we have $\Delta ABD \cong \Delta BAC$.
$(ii)$ Since $\Delta ABD \cong \Delta BAC$,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = AC$.
$(iii)$ Since $\Delta ABD \cong \Delta BAC$,their corresponding parts are equal $(CPCT)$.
Therefore,$\angle ABD = \angle BAC$.
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