Question

In a $△ABC,D$ is the mid point of side $AC$ such that $BD=\frac{1}{2}AC$. Show that $\angle ABC$ is a right angle.

Answer 1

Given that:

$BD=\frac{1}{2}AC=AD=CD$

$\Rightarrow$ Distance of $D$ is same from all vertices of $△ABC$

Hence, $D$ is circumcentre of triangle ABC.

Now $B$ would be a point on circumcircle of $△ABC$ and as angle in a semicircle is ${90}^{\circ }$, $AC$ being the diameter, we can say that:

$\angle ABC={90}^{\circ }$