In a ΔABC, D is the midpoint of side AC such that BD

Solution:

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.