Question
In 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). 
Solution
Given: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
To Prove: (i) ar(ΔACB) = ar(ΔACF)
(ii) ar(□AEDF) = ar(ABCDE).
Proof: (i) ∵ ΔACB and ΔACF are on the same base AC and between the same parallels AC and
BF. [∵ AC || BF (given)]
∴ ar(ΔACB) = ar(ΔACF)
Two triangles on the same base (or equal bases) and between the same parallels are equal in area
(ii) From (i),
ar(ΔACB) = ar(ΔACF)
⇒ ar(ΔACB) + ar(□AEDC)
= ar(ΔACF) + ar(□AEDC)
| Adding the same areas on both sides ⇒ ar(ABCDE) = ar(□AEDF)
⇒ ar(□AEDF) = ar(ABCDE).
