Question

In $△ABC$, if $\angle A={90}^o$ and $AD\perp BC$ prove that $A{D}^2=BD\times DC$.

Answer 1

REF.Image

In $\Delta BDA$, Applying Pythagoras,

$A{B}^2=B{D}^2+A{D}^2$ ________ (1)

In $\Delta ACD$, similarly

$A{C}^2=A{D}^2+D{C}^2$ _________ (2)

In $\Delta ABC$, similarly,

$B{C}^2=A{B}^2+A{C}^2$ ________ (3)

Adding (1) and (2)

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

$B{C}^2=B{D}^2+2A{D}^2+C{D}^2$ ------- (from (3))

$(BD+DC{)}^2=B{D}^2+2A{D}^2+C{D}^2$

$B{D}^2+2BD+DC.D{C}^2=B{D}^2+2A{D}^2+C{D}^2$

$\therefore BD.CD=A{D}^2$

Hence, $BD.CD=A{D}^2$