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Q3 In Fig. 6.53, ABD is a triangle right angled at A and AC $\perp$ BD. Show that $AC^2 = BC . DC .$

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Let $\angle CAB$ be x

In $\triangle ABC$ ,

$\angle CBA=180 \degree-90 \degree-x$

$\angle CBA=90 \degree-x$

Similarly,

In $\triangle CAD$ ,

$\angle CAD=90 \degree-\angle CAB$

$\angle CAD=90 \degree-x$

$\angle CDA=180 \degree-90 \degree-(90 \degree-x)=x$

In $\triangle ABC\, and\, \triangle ACD,$

$\angle CBA\, =\angle CAD$

$\angle CAB\, =\angle CDA$

$\angle ACB\, =\angle DCA$ ( Each right angle)

$\triangle ABC\, \sim \triangle ,ACD$ (By AAA)

$\frac{AC}{DC}=\frac{BC}{AC}$

$\Rightarrow AC^2=BC\times DC$

Hence proved

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