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