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

(i) Since triangles ACB and ACF are on the same base AC and between the same parallels AC and BF. Therefore ar(ACB) = ar(ACF)
(ii) Adding ar(ACDE) on both sides, we get ar(ACF) + ar(ACDE) = ar(ACB) + ar(ACDE) ⇒ ar(AEDF) = ar(ABCDE)