$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig). Show that $\Delta ABC \cong \Delta CDA$.
(N/A) Given: $l \parallel m$ and $p \parallel q$.
To prove: $\Delta ABC \cong \Delta CDA$.
Proof:
Since $l \parallel m$ and $AC$ is a transversal,
$\therefore \angle BAC = \angle DCA$ [Alternate interior angles]
Also,since $p \parallel q$ and $AC$ is a transversal,
$\therefore \angle BCA = \angle DAC$ [Alternate interior angles]
Now,in $\Delta ABC$ and $\Delta CDA$:
$\angle BAC = \angle DCA$ [Proved above]
$\angle BCA = \angle DAC$ [Proved above]
$AC = CA$ [Common side]
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\Delta ABC \cong \Delta CDA$.
To prove: $\Delta ABC \cong \Delta CDA$.
Proof:
Since $l \parallel m$ and $AC$ is a transversal,
$\therefore \angle BAC = \angle DCA$ [Alternate interior angles]
Also,since $p \parallel q$ and $AC$ is a transversal,
$\therefore \angle BCA = \angle DAC$ [Alternate interior angles]
Now,in $\Delta ABC$ and $\Delta CDA$:
$\angle BAC = \angle DCA$ [Proved above]
$\angle BCA = \angle DAC$ [Proved above]
$AC = CA$ [Common side]
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\Delta ABC \cong \Delta CDA$.
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