Question.
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) $\operatorname{ar}(\mathrm{ACB})=\operatorname{ar}(\mathrm{ACF})$
(ii) $\operatorname{ar}(\mathrm{AEDF})=\operatorname{ar}(\mathrm{ABCDE})$
(i) $\operatorname{ar}(\mathrm{ACB})=\operatorname{ar}(\mathrm{ACF})$
(ii) $\operatorname{ar}(\mathrm{AEDF})=\operatorname{ar}(\mathrm{ABCDE})$
Solution:
(i) $\triangle \mathrm{ACB}$ and $\triangle \mathrm{ACF}$ lie on the same base $\mathrm{AC}$ and are between
The same parallels AC and BF.
$\therefore$ Area $(\triangle \mathrm{ACB})=$ Area $(\triangle \mathrm{ACF})$
(ii) It can be observed that
$\Rightarrow$ Area $(\triangle \mathrm{ACB})+$ Area $(\mathrm{ACDE})=$ Area $(\mathrm{ACF})+$ Area $(\mathrm{ACDE})$
$\Rightarrow$ Area $(\triangle A C B)+$ Area $(A C D E)=$ Area $(A C F)+$ Area $(A C D E)$
$\Rightarrow$ Area $(A B C D E)=$ Area $(A E D F)$
(i) $\triangle \mathrm{ACB}$ and $\triangle \mathrm{ACF}$ lie on the same base $\mathrm{AC}$ and are between
The same parallels AC and BF.
$\therefore$ Area $(\triangle \mathrm{ACB})=$ Area $(\triangle \mathrm{ACF})$
(ii) It can be observed that
$\Rightarrow$ Area $(\triangle \mathrm{ACB})+$ Area $(\mathrm{ACDE})=$ Area $(\mathrm{ACF})+$ Area $(\mathrm{ACDE})$
$\Rightarrow$ Area $(\triangle A C B)+$ Area $(A C D E)=$ Area $(A C F)+$ Area $(A C D E)$
$\Rightarrow$ Area $(A B C D E)=$ Area $(A E D F)$
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