# $\mathrm{ABCD}$ 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.]

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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.

Updated on 10-Oct-2022 13:42:11