Line-segment $AB$ is parallel to another line-segment $CD$. $O$ is the mid-point of $AD$ (see Fig). Show that $(i)$ $\Delta AOB \cong \Delta DOC$ $(ii)$ $O$ is also the mid-point of $BC$.
$(i)$ Consider $\Delta AOB$ and $\Delta DOC$.
$\angle OAB = \angle ODC$ (Alternate interior angles as $AB \parallel CD$ and $AD$ is the transversal)
$\angle AOB = \angle DOC$ (Vertically opposite angles)
$OA = OD$ (Given,as $O$ is the mid-point of $AD$)
Therefore,by $ASA$ congruence rule,$\Delta AOB \cong \Delta DOC$.
$(ii)$ Since $\Delta AOB \cong \Delta DOC$,then by $CPCT$ (Corresponding Parts of Congruent Triangles),
$OB = OC$
Thus,$O$ is the mid-point of $BC$.
$\angle OAB = \angle ODC$ (Alternate interior angles as $AB \parallel CD$ and $AD$ is the transversal)
$\angle AOB = \angle DOC$ (Vertically opposite angles)
$OA = OD$ (Given,as $O$ is the mid-point of $AD$)
Therefore,by $ASA$ congruence rule,$\Delta AOB \cong \Delta DOC$.
$(ii)$ Since $\Delta AOB \cong \Delta DOC$,then by $CPCT$ (Corresponding Parts of Congruent Triangles),
$OB = OC$
Thus,$O$ is the mid-point of $BC$.
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$(iv)$ $AP$ is the perpendicular bisector of $BC$.
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