Question
In a $\triangle\text{ABC},\text{M}$ and $N$ are points on the sides $AB$ and $AC$ respectively such that $BM || BC.$
✓
Answer
In $\triangle\text{ABC},\angle\text{B}=\angle\text{C}$
$\therefore\text{AB}=\text{AC} ($Sides opposite to equal angle are equal$)$
Subtracting BM from both sides, we get
$AB - BM = AC - BM$
$⇒ AB - BM = AC - CN (\therefore\text{BM=CN})$
$⇒ AM = AN$
$\therefore\angle\text{AMN}=\angle\text{ANM} ($Angels opposite to equal sides are equal$)$
Now, in $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\dots(1)($Angle Sum Property of triangle$)$
Again in $\triangle\text{AMN},$
$\angle\text{A}+\angle\text{AMN}+\angle\text{ANM}=180^\circ\dots(2)($Angle Sum Property of triangle$)$
From $(1)$ and $(2),$ we get
$\angle\text{B}+\angle\text{C}=\angle\text{AMN}+\angle\text{ANM}$
$\Rightarrow2\angle\text{B}=2\angle\text{ANM}$
$\Rightarrow\angle\text{B}=\angle\text{AMN}$
Since, $\angle\text{B}$ and $\angle\text{AMN}$ are corresponding angles.
$\therefore\text{MN }||\text{ BC}$
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