Question

In fig. the side AB of parallelogram ABCD is produced to a point X such that BX=AB. Prove that DX bisects BC​

Answer 1

Answer:

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Step-by-step explanation:

ar(ΔADE)=  

2

1

​  

ar(ABCD) ............(1)

ar(ΔCEX)=  

2

1

​  

ar(ΔCBX)

(∵CE=EB)

=  

2

1

​  

ar(ΔCBA)

(∵BX=BA)

=  

2

1

​  

×  

2

1

​  

ar(ABCD)

=  

4

1

​  

ar(ABCD) ..........(2)

From (2), we can understand that

2.ar(ΔCEX)=  

2

1

​  

ar(ABCD)

=ar(ΔADE) (From (1))

(A) 2×area(ΔCEX).