Question

Question 3 (ii)
In the given figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that:
(ii) $A{C}^2=BC\times DC$

Answer 1

(ii) Let $\angle CAB=x$........(i)
In $\Delta CBA$,
$\angle CBA={180}^{\circ }-{90}^{\circ }-x$
$\angle CBA={90}^{\circ }-x$..........(ii)

$\angle BAD={90}^{\circ }$ (Given)
$\angle BAD=\angle CAB+\angle CAD$
${90}^{\circ }=x+\angle CAD$
$\angle CAD={90}^{\circ }-x$........(iii)

Similarly, in $\Delta CAD$
$\angle CDA={180}^{\circ }-{90}^{\circ }-({90}^{\circ }-x)$
$\angle CDA=x$.........(iv)

In $\Delta CBA$ and $\Delta CAD$, we have
$\angle CBA=\angle CAD$ [from (ii) and (iii)]
$\angle CAB=\angle CDA$ [from (i) and (iv)]
$\angle ACB=\angle DCA$ (Each angle is equal to ${90}^{\circ }$)
$\therefore \Delta CBA\sim \Delta CAD$ [By AAA similarity criterion]
$\Rightarrow \frac{AC}{DC}=\frac{BC}{AC}$
Therefore, $A{C}^2=BC\times DC$