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# In figure below, ABCDE is a pentagon. A line through $ \mathrm{B} $ parallel to $ \mathrm{AC} $ meets $ \mathrm{DC} $ produced at F. Show that

**(i)** $ \operatorname{ar}(\mathrm{ACB})=\operatorname{ar}(\mathrm{ACF}) $

**(ii)** $ \operatorname{ar}(\mathrm{AEDF})=\operatorname{ar}(\mathrm{ABCDE}) $

"

Given:

$ABCDE$ is a pentagon.

A line through \( \mathrm{B} \) parallel to \( \mathrm{AC} \) meets \( \mathrm{DC} \) produced at $F$.

To do:

We have to show that

(i) \( \operatorname{ar}(\mathrm{ACB})=\operatorname{ar}(\mathrm{ACF}) \)(ii) \( \operatorname{ar}(\mathrm{AEDF})=\operatorname{ar}(\mathrm{ABCDE}) \)

Solution:

$ABCDE$ is a pentagon and $BF \| AC$.

(i) $\triangle ACB$ and $\triangle ACF$ lie on the same base $AC$ and between the parallels $AC$ and $BF$.

Therefore,

$ar (\triangle ACB) = ar (\triangle ACF)$........…(i)

(ii) $ar (AEDF) = ar (AEDC + ar (\triangle ACF)$

$=ar (AEDC) + ar (\triangle ACB)$ [From (i)]

$= ar (ABCDE)$

Hence proved.

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