Question

In a right ∆ABC, ∠B = 90° and D is the mid-point of AC. Prove that $BD=\frac{1}{2}AC$.

Answer 1

AD = DC (∵ D is the midpoint)
$\angle ADB=\angle BDC$ (Altitude = BD)
BD = BD (Common side)
∴ $△ADB\cong △CDB$ (By SAS criterion)
⇒ $\angle A=\angle C$ (CPCT)
Now, we know that $\angle B=90°$.
$\Rightarrow \angle A=\angle \mathrm{ABD}=45°$ (Using sum angle property)
Similarly, $\angle BDC=90°$ (BD is altitude)
$\angle C=45°$ (Proved)
$\angle DBC=45°$
$\Rightarrow \angle ABD=45°$
$\Rightarrow BD=CDandBD=AD$ (Isosceles triangle property)
As AD+DC =AC
$$\begin{gathered} \Rightarrow BD+BD=AC \\ \Rightarrow 2BD=AC \\ \Rightarrow BD=\frac{1}{2}AC \end{gathered}$$
Hence proved.