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

Q3 (3)   In Fig. 6.53, ABD is a triangle right angled at A and AC \perp  BD. Show that
              AD^2 = BD . CD .

             

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In \triangle ACD\, and\, \triangle ABD,

\angle DCA\, =\angle DAB\, \, \, \, \, \, \, \, (Each 90 \degree)

\angle CDA\, =\angle ADB       (common )

\triangle ACD\, \sim \triangle ABD                (By AA)

\Rightarrow \frac{CD}{AD}=\frac{AD}{BD}

\Rightarrow AD^2=BD\times CD

Hence prooved

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