i) ΔACB and ΔACF are lying on the same base AC and are existing between the same parallels AC and BF.
According to Theorem 9.2: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Therefore, ar (ΔACB) = ar (ΔACF)
ii) It can be observed that
ar (ΔACB) = ar (ΔACF)
Adding area (ACDE) on both the sides.
Area (ΔACB) + Area (ACDE) = Area (ΔACF) + Area (ACDE)
Area (ABCDE) = Area (AEDF)