Question

In triangle $ABC,AD$ is perpendicular to side $BC$ and $A{D}^2=BD\times DC$.
Show that $\angle$ $BAC={90}^o$ ?

Answer 1

$Given:IntriangleABC,ADisperpendiculartoBCandA{D}^2=BD.DC$

$Toprove:BAC=90°$

$Proof:Inrighttriangles\Delta ADBand\Delta ADC$

$So,Pythagorastheoremshouldbeapply,$

$Thenwehave,$

$A{B}^2=A{D}^2+B{D}^2----------\left(1\right)$

$A{C}^2=A{D}^2+D{C}^2---------\left(2\right)$

$A{B}^2+A{C}^2=2A{D}^2+B{D}^2+D{C}^2$

$=2BD.CD+B{D}^2+C{D}^2,[\because givenA{D}^2=BD.CD]$

$=(BD+CD{)}^2=B{C}^2$

$ThusintriangleABCwehave,A{B}^2+A{C}^2=B{C}^2$

$hencetriangleABCisarighttriangle$

$rightangledhence\angle BAC=90°$