In Fig. below, CD || AE and CY || BA.

(i) Name a triangle equal in area of ΔCBX

(ii) Prove that ar (Δ ZDE) = ar (ΔCZA)

(iii) Prove that ar (BCZY) = ar (Δ EDZ)

#### Solution

Since, ΔBCA and ΔBYAare on the same base BA and between same parallels BA and CY Then area (ΔBCA) = ar (BYA)

⇒ ar (Δ CBX) + ar (Δ BXA) = ar (Δ BXA) + ar (ΔAXY)

⇒ ar (Δ CBX) = ar (Δ AXY) ........ (1)

Since, ΔACE and ΔADE are on the same base AE and between same parallels CD and AE

Then, ar (ΔACE) ar (ΔADE)

⇒ (ΔCLA) + ar (ΔAZE) = ar (ΔAZE) + ar (ΔDZE)

⇒ ar (ΔCZA) = (ΔDZE) ........ (2)

Since ΔCBY and ΔCAY are on the same base CY and between same parallels

BA and CY

Then ar (ΔCBY) ar (ΔCAY )

Adding ar (ΔCYG) on both sides, we get

⇒ ar (ΔCBX) + ar (ΔCYZ) = ar (ΔCAY) + ar (ΔCYZ)

⇒ ar (BCZX) = ar (Δ CZA) ......... (3)

Compare equation (2) and (3)

ar (BCZY) = ar (ΔDZE)