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)