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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