Question

In Fig., $ABD$ is a triangle right angled at $A$ and $AC$ $\perp$ BD. Show that:
(i) $A{B}^2=BC.BD$
(ii) $A{C}^2=BC.DC$
(iii) $A{D}^2=BD.CD$

Answer 1

In $△DCA$,
By Pythagoras theorem,
$(DA{)}^2=(CA{)}^2+(CD{)}^2.......(1)$
In $△ABD$,
By Pythagoras theorem,
$(BD{)}^2=(AB{)}^2+(AD{)}^2........(2)$
In $△ABC$,
(i)
By Pythagoras theorem,
$(AB{)}^2=(AC{)}^2+(BC{)}^2.......(3)$
$\therefore$ $(AB{)}^2=(AD{)}^2-(CD{)}^2+(BC{)}^2$ (From 1 and 3)
$\therefore$ $(AB{)}^2=(BD{)}^2-(AB{)}^2-(CD{)}^2+(BC{)}^2$
$\therefore$ $2(AB{)}^2=(BD{)}^2-(BD-BC{)}^2+(BC{)}^2$
$\therefore$ $(AB{)}^2=BC.BD$
(ii)
Adding $\left(1\right)$ and $\left(3\right),$
$\therefore$ $(AB{)}^2+(AD{)}^2=2(AC{)}^2+(CD{)}^2+(BC{)}^2$
$\therefore$ $(BD{)}^2-(AD{)}^2+(AD{)}^2=2(AC{)}^2+(CD{)}^2+(BC{)}^2$ (From 2)
$\therefore$ $(BC+CD{)}^2=2(AC{)}^2+(CD{)}^2+(BC{)}^2$
Hence, Solving the above equation we get,
$(AC{)}^2=BC.DC$
(iII)
Subtracting $\left(1\right)$ and $\left(2\right)$ we get,
$2(AD{)}^2-(BD{)}^2=(CA{)}^2+(CD{)}^2-(AB{)}^2$
$\therefore$ $2(AD{)}^2-(BD{)}^2=(CA{)}^2+(CD{)}^2-(AC{)}^2-(BC{)}^2$
$\therefore$ $2(AD{)}^2-(BD{)}^2=(BD-BC{)}^2-(BC{)}^2$
Hence, Solving the above equation we get,
$(AD{)}^2=BD.DC$