In the figure, $CD \parallel AE$ and $CY \parallel BA$.
Name a triangle equal in area of $\triangle CBX$.
"
Given:
$CD \parallel AE$ and $CY \parallel BA$.
To do:
We have to name a triangle equal in area of $\triangle CBX$.
Solution:
$\triangle CBX$ and $\triangle CYX$ are on the same base $BY$ and between same parallels.
Therefore,
$ar(\triangle CBX) = ar(\triangle CYX)$.
The triangle equal in area of triangle $CBX$ is $\triangle CYX$.
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