Question

Question 13
In a triangle ABC, D is the mid-point of side AC such that BD = $\frac{1}{2}$ AC. Show that $\angle ABC={90}^{\circ }$.

Answer 1

We have, $BD=\frac{1}{2}AC...\left(i\right)$
D is the mid-point of AC.
$\therefore AD=CD=\frac{1}{2}AC...\left(ii\right)$
From Eqs. (i) and (ii),
AD = CD = BD
In $\Delta ADB,AD=BD\left[provedabove\right]$
$\therefore \angle ABD=\angle BAD...\left(iii\right)$
[angles opposite to equal sides are equal]
In $\Delta DBC,BD=CD\left[provedabove\right]\therefore \angle BCD=\angle CBD...\left(iv\right)$
[angles opposite to equal sides are equal]
In $\Delta ABC,\angle ABC+\angle BAC+\angle ACB={180}^{\circ }$ [by angle sum property of a triangle]
$\Rightarrow \angle ABC+\angle BAD+\angle DCB={180}^{\circ }\Rightarrow \angle ABC+\angle ABD+\angle CBD={180}^{\circ }fromEqs.\left(iii\right)and\left(iv\right)]\Rightarrow \angle ABC+\angle ABC={180}^{\circ }\Rightarrow \angle ABC={90}^{\circ }$