Question

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that(i) ar(ΔACB)=ar(ΔACF)(ii) ar(AEDF)=ar(ABCDE)

Answer 1

Two Triangles on the same base and between the same parallels are equal in area.

Given:ABCDE is a Pentagon & BF||AC.

To show:(i) ar (ACB) = ar (ACF) ii) ar (AEDF) = ar (ABCDE)

Proof:

i) △ACB and △ACF lie on the same base AC and between the same parallels AC and BF

.∴ ar(△ACB) = ar(△ ACF)

(ii)ar(△ACB) = ar(△ACF)

ar(△ACB)+ar(△ACDE) =ar(△ACF) + ar(△ACDE)

[ On adding ar(△ACDE) on both sides]

ar(ABCDE) = ar(△AEDF)

Hope this will help you...

Answer 2

Hi friend..
See the attached files

I hope it will help you
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