$A B C D$ is a parallelogram such that $L$ is mid-point of $B C$, then $\mathbf{A L}$ is equal to

$A B C D$ is a parallelogram such that $L$ is mid-point of $B C$, then $\mathbf{A L}$ is equal to
  1. $\mathrm{DC}+\frac{1}{2} \mathrm{AD}$
  2. $\frac{1}{2} A D+B C$
  3. $\frac{1}{2} \mathrm{AD}+\mathrm{DL}$
  4. $\frac{1}{2} \mathrm{AD}+\mathrm{BL}$

Solution

$A B C D$ is a parallelogram such that $L$ is mid-point on $B C$
Now for $\mathbf{A L}$ Consider, $\triangle A B L$ $ \mathbf{A B}+\mathbf{B L}=\mathbf{A L} $ or $ \begin{aligned} & \mathbf{A L}=\mathbf{A B}+\mathbf{B L} [\mathbf{B C}=\mathbf{A D}]\\ & =\mathbf{D C}+\frac{1}{2}(\mathbf{B C})=\mathbf{D C}+\frac{1}{2} \mathbf{A D} \end{aligned} $ $[\because A B C D$ is a parallelogram, $\mathbf{A B}=\mathbf{D C}$ and $L$ is mid-point of $B C$, then $\left.\mathbf{B L}=\frac{1}{2} \mathbf{B C}\right]$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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