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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