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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`.

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In `DBAC` and `DeltaADC`
`/_BAC~=/_ADC`..(Given)
`/_ACB~=/_DCA`………(Common angle)
`:.DeltaBAC~DeltaADC` ………(AA test of similarity)
`:.(CA)/(CD)=(CB)/(CA)` ………..(Corresponding sides of similar triangles are in proportion)
`:.CA^(2)=CBxxCD.`
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NAVNEET PUBLICATION - MAHARASHTRA BOARD-SIMILARITY-EXAMPLE TYPE
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