Question

In the figure, BED = BDE and E divides BC in the ratio 2 : 1. Prove that AF x BE = 2 AD
CF.
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Answer 1

Step-by-step explanation:

Through C draw CG || FD.

Now, In △AFD,

CG || FD

⇒ AC = AG ............[BPT]

CF GD

⇒ AC + 1 = AG + 1

CF GD

⇒ AC + CF = AG + GD

CF GD

⇒ AF = AD ------ ( 1 )

CF GD

In △BCG,

Since, DE || GC, then

⇒ BE = BD ...........[BPT]

EC GD

⇒ 2 = BD ..........[ As, BE:EC=2:1 ]

1 GD

⇒ BD = 2GD ----- ( 2 )

In △BDE,

⇒ ∠BED = ∠BDE ...........[ Given ]

⇒ BD = BE ...........[Sides opposite to equal angles are equal] ----- ( 3 )

From ( 2 ) and ( 3 ), we get

⇒ 2GD = BE

⇒ GD = BE ----- ( 4 )

2

Substituting the value of GD from ( 4 ) in ( 1 ), we get

⇒ AF = AD

CF BE

2

⇒ AF = 2AD

CF BE

⇒ AF × BE = 2AD × CF

HENCE PROVED!!!