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

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

              

Answers (1)
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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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