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Q13 D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$ . Show
that $CA^2 = CB.CD.$

Answers (1)

S seema garhwal

In, $\triangle ADC \, \, and\, \, \triangle BAC,$

$\angle ADC = \angle BAC$ ( given )

$\angle ACD = \angle BCA$ (common )

$\triangle ADC \, \, \sim \, \, \triangle BAC,$ ( By AA rule)

$\frac{CA}{CB}=\frac{CD}{CA}$ ( corresponding sides of similar triangles )

$\Rightarrow CA^2=CB\times CD$

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