Question

In the fig angle BED=BDE and E divides BC in the ratio 2:1.prove that AF×BE=2AD ×CF

Answer 1

Answer:

AF×BE=2AD×CF

Step-by-step explanation:

Step : 1 Through C draw CG∥FD.

Now, In△AFD,

CG∥FD

⇒  $\frac{\mathrm{AC}}{\mathrm{CF}}=\frac{\mathrm{AG}}{\mathrm{GD}}$            [ BPT ]

$\Rightarrow \frac{\mathrm{AC}}{\mathrm{CF}}+1=\frac{\mathrm{AG}}{\mathrm{GD}}+1$

$\Rightarrow \frac{\mathrm{AC}+\mathrm{CF}}{\mathrm{CF}}=\frac{\mathrm{AG}+\mathrm{GD}}{\mathrm{GD}}$

$\Rightarrow \frac{\mathrm{AF}}{\mathrm{CF}}=\frac{\mathrm{AD}}{\mathrm{GD}}$               ------ ( 1 )

Step : 2  In △BCG,

Since, DE∥GC, then

$\Rightarrow \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BD}}{\mathrm{GD}}$
$\Rightarrow \frac{2}{1}=\frac{\mathrm{BD}}{\mathrm{GD}} \quad[\mathrm{As}, \mathrm{BE}: \mathrm{EC}=2: 1]$

$\Rightarrow \mathrm{BD}=2 \mathrm{GD}$

Step : 3 In △BDE,

$\Rightarrow \angle \mathrm{BED}=\angle \mathrm{BDE}$

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

From ( 2 ) and ( 3 ), we get

⇒  2GD=BE

$\Rightarrow \mathrm{GD}=\frac{\mathrm{BE}}{2}$               ----- ( 4 )

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

$\Rightarrow \frac{\mathrm{AF}}{\mathrm{CF}}=\frac{\mathrm{AD}}{\frac{\mathrm{BE}}{2}}$

$\Rightarrow \frac{\mathrm{AF}}{\mathrm{CF}}=\frac{2 \mathrm{AD}}{\mathrm{BE}}$

⇒  AF×BE=2AD×CF

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