Question
In $\triangle\text{ABC,D}$ is the midpoint of $BC$. If $\text{DL}\perp\text{AB}$ and $\text{DM}\perp\text{AC}$ such that $DL = DM$, prove that $AB = AC$. 
✓
Answer
Given: In a $\triangle\text{ABC,D}$ is the midpoint of $BC$ and $\text{DL}\perp\text{AB}$ and $\text{DM}\perp\text{AC}.$
Also, $DL = DM$
To prove: $\text{AB = AC}$
proof: In right angle triangles $\triangle\text{BLD}$ and $\triangle\text{CMD}$
$\angle\text{BLD}=\angle\text{CMD}=90^{\circ}$
$\text{Hypt.BD = Hypt.CD}$ [Given] $\text{DL = DM}$ [Given]
Thus, by Right Angle-Hypotenuse-Side criterion of congruence,
we have $\triangle\text{BLD}=\triangle\text{CMD}$ [By $RHS$]
The corresponding parts of the congruent triangle are equal.
$\therefore\angle\text{ABD}=\angle\text{ACD}$ $[C.P.C.T.]$ In
$\triangle\text{ABC,}$we have $\angle\text{ABD}=\angle\text{ACD}$
$\Rightarrow\text{AB = AC}$ [$\therefore$ side oposite to equal angles are equal]
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