Question
In a parallelogram $ABCD$, points $M$ and $N$ have been taken on opposite sides $AB$ and $CD$ respectively such that $AM = CN.$ Show that $AC$ and $MN$ bisect each other.
✓
Answer
In $\triangle\text{AMO}$ and $\triangle\text{CNO}$
$\angle\text{MAO = }\angle\text{NCO}$ ($AB \| CD$, alternate angles)
$\text{AM = CN}$ (given)
$\angle\text{AOM}=\angle\text{CON}$ (vertically opposite angles)
$\therefore\triangle\text{AMO}\cong\triangle\text{CNO}$ (by $ASA$ congruence criterion)
$\Rightarrow\text{AO = CO}$ and $\text{MO = NO} (CP.C.T.)$
$\Rightarrow\text{AC}$ and $\text{MN}$ bisect each other.
$\angle\text{MAO = }\angle\text{NCO}$ ($AB \| CD$, alternate angles)
$\text{AM = CN}$ (given)
$\angle\text{AOM}=\angle\text{CON}$ (vertically opposite angles)
$\therefore\triangle\text{AMO}\cong\triangle\text{CNO}$ (by $ASA$ congruence criterion)
$\Rightarrow\text{AO = CO}$ and $\text{MO = NO} (CP.C.T.)$
$\Rightarrow\text{AC}$ and $\text{MN}$ bisect each other.
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