Question

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

Answer 1

Given: $D$ is the midpoint of side $AC$ such that $BD=\frac{1}{2}AC.$

$AD=CD=\frac{1}{2}AC$ [$\because D$ is the midpoint of $AC$]

$BD=\frac{1}{2}AC$ [Given]

$\therefore AD=CD=BD$

In $△ADB,AD=BD$

$\Rightarrow \angle DAB=\angle DBA=\angle x$ (angles opposite equal sides)

In $△BDC,$ we have

$\Rightarrow BD=CD$ [from above]

$\Rightarrow \angle DBC=\angle DCB=\angle y$ (angles opposite equal sides)

In $\Delta ABC,$ we have

$\angle ABC+\angle BCA+\angle CAB={180}^{\circ }$ [angle sum property]

$\Rightarrow (\angle x+\angle y)+\angle y+\angle x={180}^{\circ }$

$\Rightarrow \angle x+\angle y={90}^{\circ }$

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