$ \mathrm{ABCD} $ is a trapezium with $ \mathrm{AB} \| \mathrm{DC} $. A line parallel to $ \mathrm{AC} $ intersects $ \mathrm{AB} $ at $ \mathrm{X} $ and $ \mathrm{BC} $ at Y. Prove that ar $ (\mathrm{ADX})=\operatorname{ar}(\mathrm{ACY}) $.
[Hint: Join CX.]

Given:

\( \mathrm{ABCD} \) is a trapezium with \( \mathrm{AB} \| \mathrm{DC} \). A line parallel to \( \mathrm{AC} \) intersects \( \mathrm{AB} \) at \( \mathrm{X} \) and \( \mathrm{BC} \) at Y. 

To do:

We have to prove that ar \( (\mathrm{ADX})=\operatorname{ar}(\mathrm{ACY}) \).

Solution:

$AC \| XY$

Join $CX, AY$ and $DX$

$\triangle ADX$ and $\triangle ACX$ lie on the same base $AX$ and between the parallels $AB$ and $DC$.

Therefore,

$ar(\triangle ADX) = ar(\triangle ACX)$.....…(i)

$\triangle ACX$ and $\triangle ACY$ lie on the same base $AC$ and between the parallels $AC$ and $XY$.

Therefore,

$ar(\triangle ACX) = ar(\triangle ACY)$.......…(ii)

From (i) and (ii), we get,

$ar(\triangle ADX) =ar(\triangle ACX)= ar(\triangle ACY)$

This implies,

$ar(\triangle ADX) = ar(\triangle ACY)$

Hence proved.