In the figure,$OA = OB$ and $OD = OC$. Show that
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.
(N/A) $(i)$ In $\Delta AOD$ and $\Delta BOC$:
$OA = OB$ (Given)
$OD = OC$ (Given)
Also,$\angle AOD = \angle BOC$ (Vertically opposite angles).
Therefore,by the $SAS$ congruence rule,$\Delta AOD \cong \Delta BOC$.
$(ii)$ Since $\Delta AOD \cong \Delta BOC$,their corresponding parts are equal $(CPCT)$.
Thus,$\angle OAD = \angle OBC$.
These are alternate interior angles formed by the transversal $AB$ intersecting lines $AD$ and $BC$.
Since the alternate interior angles are equal,$AD \parallel BC$.
$OA = OB$ (Given)
$OD = OC$ (Given)
Also,$\angle AOD = \angle BOC$ (Vertically opposite angles).
Therefore,by the $SAS$ congruence rule,$\Delta AOD \cong \Delta BOC$.
$(ii)$ Since $\Delta AOD \cong \Delta BOC$,their corresponding parts are equal $(CPCT)$.
Thus,$\angle OAD = \angle OBC$.
These are alternate interior angles formed by the transversal $AB$ intersecting lines $AD$ and $BC$.
Since the alternate interior angles are equal,$AD \parallel BC$.
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