Question

In $\Delta ABC,AD\perp BC$ such that $A{D}^2=BD\times CD.$ Prove that $\Delta ABC$ is right angle at $A?$

Answer 1

$\begin{array}{c}Inright\Delta ADBand\Delta ADC\\ A{B}^2=A{D}^2+B{D}^2-----\left(1\right)\\ A{C}^2=A{D}^2+D{C}^2------\left(2\right)\\ Nownyaddingthem\\ A{B}^2+A{C}^2=2A{D}^2+B{D}^2+D{C}^2\\ =2BD.CD+B^2+C{D}^2\left[A{D}^2=BD.CDgiven\right]\\ ={\left(BD+CD\right)}^2\\ =B{C}^2\\ Thus,in\Delta ABCwehave\\ A{B}^2+A{C}^2=B{C}^2\\ So,\Delta ABCisarighttrianglerightangleatA.\\ \therefore \angle BAC={90}^{\circ }.\end{array}$