In the figure, $CD \parallel AE$ and $CY \parallel BA$.
Prove that $ar(\triangle ZDE) = ar(\triangle CZA)$.
"
Given:
$CD \parallel AE$ and $CY \parallel BA$.
To do:
We have to prove that $ar(\triangle ZDE) = ar(\triangle CZA)$.
Solution:
$\triangle ADE$ and $\triangle ACE$ are on the same base $AE$ and between the same parallels.
Therefore,
$ar(\triangle ADE) = ar(\triangle ACE)$
Subtracting $ar(\triangle AZE)$ from both sides, we get,
$ar(\triangle ADE) - ar(\triangle AZE) = ar(\triangle ACE) - ar(\triangle AZE)$
$ar(\triangle ZDE) = ar(\triangle ACZ)$
$ar(\triangle ZDE) = ar(\triangle CZA)$
Hence proved.
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