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Parallelogram

ABCD is a par...

Question

ABCD is a parallelograme if L and M are the points of BC and CD respectively, then find $¯ A L$ and $¯ A M$ interms of $¯ A B$ and $¯ A D$

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Solution

Given that,

$A B C D$ is a parallelogram an $L, M$ are the midpoints of $B C$ and $C D$ respectively.

Now, In $Δ A L C$ , Using law of addition.

$- - \to A L + - - \to L C = - - \to A C$

But $L$ is the mid point of $- - \to B C$

$- - \to L C = \frac{1}{2} - - \to B C$

$- - \to A L + \frac{1}{2} - - \to B C = - - \to A C$

$- - \to A L = - - \to A C - \frac{1}{2} - - \to B C . . . . . . . (2)$

But, since $M$ is the mid point of $- - \to D C$

Now,

$- - \to M C = \frac{1}{2} - - \to D C$

$\Rightarrow - - \to A M + \frac{1}{2} - - \to D C = - - \to A C$

$\Rightarrow - - \to A M = - - \to A C - \frac{1}{2} - - \to D C . . . . . . . (2)$

On adding (1) and (2) to, we get,

$- - \to A L + - - \to A M = - - \to A C - \frac{1}{2} - - \to B C + - - \to A C - \frac{1}{2} - - \to D C$

$= 2 - - \to A C - \frac{1}{2} (- - \to B C + - - \to D C)$

$N o w,$

$- - \to D C = - - \to A B (A B C D is a paralellogram)$

$\Rightarrow - - \to A L + - - \to A M = 2 - - \to A C - \frac{1}{2} (- - \to B C + - - \to A B)$

$- - \to A B + - - \to B C = - - \to A C$

$\Rightarrow - - \to A L + - - \to A M = 2 - - \to A C - \frac{1}{2} - - \to A C$

$\Rightarrow - - \to A L + - - \to A M = \frac{3}{2} - - \to A C$

$\Rightarrow - - \to A L + - - \to A M = \frac{3}{2} (- - \to B C + - - \to A B)$

$Now, - - \to B C = - - \to A D (It is a paralellogram)$

$\Rightarrow - - \to A L + - - \to A M = \frac{3}{2} (- - \to A D + - - \to A B)$
Hence proved.

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