Question

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

Answer 1

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

To prove: ∠ABC is a right angle.

Proof:

$$\begin{gathered} \mathrm{In}∆ADB, \\ AD=BD\left(\mathrm{Given},BD=\frac{1}{2}AC\right) \\ \Rightarrow \angle DAB=\angle DBA=x\left(\mathrm{Let}\right)\left(\mathrm{Angles}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Similarly},\mathrm{in}∆DCB, \\ BD=CD\left(\mathrm{Given}\right) \\ \Rightarrow \angle DBC=\angle DCB=y\left(\mathrm{Let}\right) \end{gathered}$$

$\mathrm{In}∆ABC,$

$$\begin{gathered} \angle ABC+\angle BCA+\angle CAB=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow x+x+y+y=180° \\ \Rightarrow 2(x+y)=180° \\ \Rightarrow x+y=90° \\ \Rightarrow \angle ABC=90° \end{gathered}$$

Hence, ∠ABC is a right angle.