In Fig. 9.27, 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)

Solution:

1. △ACB and △ACF lie on the same base AC and between the same parallels AC and BF.

∴ar(△ACB) = ar(△ ACF)

1. ar(△ACB) = ar(△ACF)

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

⇒  ar(ABCDE) = ar(AEDF)