In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that: $a r e a (A E D F) = a r e a (A B C D E)$

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Solution

In the figure, ABCDE is a pentagon. A line through B parallel to AC meet DC produced at F.

Then AC parallel to BF

Given ABCDE is a pentagon then AB parallel to DC and DC extended up to F

Then AB parallel to CF

The $Δ A C B$ and $A C F$ are same base AC and between two parallel line AC and BF

$∴ a r e a (Δ A C B) = a r e a (Δ A C F)$

Add an area of (AEDC) on both sides we get

$\Rightarrow a r e a (Δ A C B) + a r e a (A E D C) = a r e a (Δ A C F) + a r e a (A E D C)$

$\Rightarrow a r e a (A B C D E) = a r e a (A E D F)$ $[h e n c e p r o v e d]$

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In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that: area(AEDF)=area(ABCDE)

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that: $a r e a (A E D F) = a r e a (A B C D E)$