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In the given figure, if $LM \| CB$ and $LN \| CD$.
Prove that $ \frac{\mathbf{A M}}{\mathbf{A B}}=\frac{\mathbf{A N}}{\mathbf{A D}} $
"
Given:
$LM \| CB$ and $LN \| CD$.
To do:
We have to prove that \( \frac{\mathbf{A M}}{\mathbf{A B}}=\frac{\mathbf{A N}}{\mathbf{A D}} \)
Solution:
We know that,
If a line divides two sides of a triangle proportionally, then it is parallel to the third side.
In $\triangle ABC, LM \| CB$,
This implies,
$\frac{AM}{AB}=\frac{AL}{AC}$.........(i)
In $\triangle ADC, LN \| CD$,
This implies,
$\frac{AN}{AD}=\frac{AL}{AC}$.........(ii)
From (i) and (ii), we get,
$\frac{AM}{AB}=\frac{AN}{AD}$
Hence proved.
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