In figure, ABCDE is a pentagon. A line through B paralkl to AC meets E DC produced at F. Show that(i\") ar(ACBJ = ar(ACFJ(ii) ar(AEDFJ = ar(ABCDEJ

Triangles ABC and AFC are on the same base AC and between the same parallels AC and BF. (i) :. cu(ACB) = ar(ACF). (ii) Adding ar(AEDC) to both sides, we t cu(ACB) + cu(AEDC) = cu(ACF) + cu(AEDC) cu(ABCDE) = ar(AFCDE) or ar(ABCDE) = ar(AEDF)