In a $\triangle A B C, D$ is the midpoint of side $A C$ such that $B D=\frac{1}{2} A C$. Show that $\angle A B C$ is a right angle.
Given: In $\triangle A B C, D$ is the midpoint of side $A C$ such that $B D=\frac{1}{2} A C$.
To prove: $\angle A B C$ is a right angle.
Proof:
In $\Delta A D B$,
$A D=B D \quad\left(\right.$ Given, $\left.B D=\frac{1}{2} A C\right)$
$\Rightarrow \angle D A B=\angle D B A=x$ (Let) $\quad$ (Angles opposite to equal sides are equal)
Similarly, in $\Delta D C B$
$B D=C D$ (Given)
$\Rightarrow \angle D B C=\angle D C B=y$ (Let)
In $\Delta A B C$,
$\angle A B C+\angle B C A+\angle C A B=180^{\circ}$ (Angle sum property)
$\Rightarrow x+x+y+y=180^{\circ}$
$\Rightarrow 2(x+y)=180^{\circ}$
$\Rightarrow x+y=90^{\circ}$
$\Rightarrow \angle A B C=90^{\circ}$
Hence, $\angle A B C$ is a right angle.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.