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

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.

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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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