In the figure,two lines $AB$ and $CD$ intersect each other at the point $O$ such that $BC \parallel DA$ and $BC = DA$. Show that $O$ is the midpoint of both the line segments $AB$ and $CD$.
(N/A) Given: $BC \parallel AD$ and $BC = DA$.
Consider $\triangle BOC$ and $\triangle AOD$:
$1$. $\angle CBO = \angle DAO$ (Alternate interior angles,since $BC \parallel AD$)
$2$. $\angle BCO = \angle ADO$ (Alternate interior angles,since $BC \parallel AD$)
$3$. $BC = DA$ (Given)
By the $ASA$ (Angle-Side-Angle) congruence criterion,$\triangle BOC \cong \triangle AOD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
$OB = OA$ and $OC = OD$.
Therefore,$O$ is the midpoint of both the line segments $AB$ and $CD$.
Consider $\triangle BOC$ and $\triangle AOD$:
$1$. $\angle CBO = \angle DAO$ (Alternate interior angles,since $BC \parallel AD$)
$2$. $\angle BCO = \angle ADO$ (Alternate interior angles,since $BC \parallel AD$)
$3$. $BC = DA$ (Given)
By the $ASA$ (Angle-Side-Angle) congruence criterion,$\triangle BOC \cong \triangle AOD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
$OB = OA$ and $OC = OD$.
Therefore,$O$ is the midpoint of both the line segments $AB$ and $CD$.
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