detailed solution

Correct option is

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)

Hence proved.

detailed solution

Correct answer is

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)

Hence proved.