In the figure, $CD \parallel AE$ and $CY \parallel BA$.
Prove that $ar(\triangle CZY) = ar(\triangle EDZ)$.
"
Given:
$CD \parallel AE$ and $CY \parallel BA$.
To do:
We have to prove that $ar(\triangle CZY) = ar(\triangle EDZ)$.
Solution:
$ACY$ and $BCY$ are on the same base $CY$ and between the same parallels
Therefore,
$ar(\triangle ACY) = ar(\triangle BCY)$
$ar(\triangle ACZ) = ar(\triangle ZDE)$
$ar(\triangle ACY) + ar(\triangle CYZ) = ar(\triangle EDZ)$
$ar(\triangle BCY) + ar(\triangle CYZ) = ar(\triangle EDZ)$
$ar quad. (BCZY) = ar(EDZ)$
Hence proved.
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