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In the figure in DeltaABC, point D on si...

In the figure in `DeltaABC,` point D on side BC is such that `/_BAC=/_ADC`.
Prove that `CA^(2)=CBxxCD`.

Text Solution

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In `DeltaBAC` and `DeltaADC`
`/_BAC~=/_ADC`……(Given)
`/_ACB~=/_DCA`…….(Common angle)
`:.DeltaBAC~Delta ADC`…..("AA" test of similarity)
`:.(CA)/(CD)=(CB)/(CA)`……(Corrresponding sides of similar triangles are proportional)
`:.CA^(2)=CBxxCD`.
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