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) \( \operatorname{ar}(\mathrm{AEDF})=\operatorname{ar}(\mathrm{ABCDE}) \)
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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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