Question

In $△ABC$, $AD=9,DC=4$ and $BD=6$. Prove that $BD\perp AC$

Answer 1

$AD=9$

$DC=4$

$BD=6$

Let $\overline{BD}$ is perpendicular to $\overline{AC}$

$\therefore$ In $△ABD$

${AB}^2={AD}^2+{BD}^2$

$\Rightarrow {AB}^2=81+36$

$\Rightarrow AB=\sqrt{117}=3\sqrt{18}$

$\therefore$ In $△BDC$

${BC}^2={BD}^2+{CD}^2$

$\Rightarrow {BC}^2=36+16$

$\Rightarrow {BC}^2=52$

$\Rightarrow BC=\sqrt{52}=2\sqrt{13}$

Now, $\because △ABC$ is a right-angle triangle.

$\therefore {AB}^2+{BC}^2=117+52$

$=169$

${\left(13\right)}^2={\left(AC\right)}^2$

$\therefore$ our assumption is correct.

$\therefore BD\perp AC\left(proved\right)$