14. In the parallelogram ABCD, the side AB is produced
to X, so that BX = AB. The line DX cuts BC at E.
Prove that
(i) DBXC is a parallelogram;
(ii) area AED = twice area CEX.
Answers
DBXC is a parallelogram Area of ΔAED =2Area of ΔCXE
Step-by-step explanation:
BX = AB
AB = DC
=> BX = DC
AB is extended => BX ║ DC
ΔCED & ΔBEX
BX = CD
∠CED = ∠BEX ( opposite angles)
∠DCE = ∠XBE ( DC ║ BX)
=> ΔCED ≅ ΔBEX
BE = EC
DE = EX
now
ΔBED & ΔCEX
BE = CE
DE = XE
∠BEX = ∠CEX
=> ΔBED ≅ ΔCEX
BD = CX
now in DBXC
DB = CX
BX = CD & BX ║DC
=> DBXC is a parallelogram
ΔAXD ≈ ΔBXE
BE = BC/2 = AD/2
=> Area of ΔAXD = 4 * Area of ΔBXE
Area of ΔAED = Area of ΔAXD - Area of ΔAEX
=> Area of ΔAED = 4 * Area of ΔBXE - Area of ΔAEX
now in Δ AEX B is mid point of AX ( AB = BX)
=> area of Δ AEX = 2Area of ΔBXE
=> Area of ΔAED = 4 * Area of ΔBXE - 2Area of ΔBXE
=> Area of ΔAED = 2Area of ΔBXE
Area of ΔBXE = Area of ΔCXE ( as E is mid point)
=> Area of ΔAED =2Area of ΔCXE
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IN A PARALLELOGRAM ABCD
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