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# In figure below, ar $ (\mathrm{DRC})= $ ar $ (\mathrm{DPC}) $ and ar $ (\mathrm{BDP})=\operatorname{ar}(\mathrm{ARC}) $. Show that both the quadrilaterals $ \mathrm{ABCD} $ and $ \mathrm{DCPR} $ are trapeziums.

"

Given:

$ar(\mathrm{DRC})=ar(\mathrm{DPC})$ and $ar(\mathrm{BDP})=\operatorname{ar}(\mathrm{ARC})$

To do:

We have to show that both the quadrilaterals \( \mathrm{ABCD} \) and \( \mathrm{DCPR} \) are trapeziums.

Solution:

$ar (\triangle DPC) = ar(\triangle DRC)$.....……(i)

$ar(\triangle BDP) = ar(\triangle ARC)$……...(ii)

Subtracting (i) from (ii), we get,

$ar(\triangle BDP) - ar(\triangle DPC) = ar(\triangle ARC) - ar(\triangle DRC)$

$ar(\triangle BDC) = ar(\triangle ADC)$

Triangles $BDC$ and $ADC$ are on the same base $DC$.

Therefore,

$DC \| AB$

This implies,

$ABCD$ is a trapezium.

Similarly,

$ar(\triangle DRC) = ar(\triangle DPC)$

Triangles $DRC$ and $DPC$ have the same base $DC$.

Therefore,

$RP \| DC$

This implies,

$DCPR$ is a trapezium.

Hence proved.

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