Question

In the figure given below$,\angle BAC={90}^0$ and $AD\perp BC.$ then $A{D}^2=BD\times DC.$
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In $△ABD$,
$A{B}^2=B{D}^2+A{D}^2$ (Pythagoras theorem)
In $△ACD$,
$A{C}^2=C{D}^2+A{D}^2$ (Pythagoras theorem)
Add both the equations,
$A{B}^2+A{C}^2=B{D}^2+A{D}^2+C{D}^2+A{D}^2$
$B{C}^2=B{D}^2+A{D}^2+2BD\times CD-2BD\times CD+2A{D}^2$ (Apply pythagoras theorem to triangle ABC)
$B{C}^2=(BD+DC{)}^2-2BD\times CD+2A{D}^2$
$B{C}^2=B{C}^2-2BD\times CD+2A{D}^2$
$A{D}^2=BD\times CD$