Question
Given: Parallelogram $ABCD$ in which diagonals $AC$ and $BD$ intersect at $M$. Prove: $M$ is the mid$-$point of $LN$.
✓
Answer
Proof: Diagonals of gm bisect each other.
$MD = MB$
Also $\angle ADB = \angle DBN ($Alternate$ \angle s)$
$\angle DML = \angle BMN ($Vert. opp. $\angle s)$
$\triangle DML = \triangle BMN$
$LM = MN$
$M$ is mid$-$point of $LN$.
Hence proved.
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(a) $\frac{x}{2}+\frac{3 x}{4}=x+9$ $\quad$ (b) $\frac{x}{2}+\frac{3 x}{8}=x+9$ $\quad$ (c) $\frac{x}{2}+\frac{3 x}{4}=x-9$ $\quad$ (d) $\frac{x}{2}+\frac{3 x}{4}=x-9$
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