Math, asked by Anonymous, 1 year ago

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

Answered by amitnrw
13

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

Learn More :

IN A PARALLELOGRAM ABCD

https://brainly.in/question/11588611

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