A) \[AB\times AC\]
B) \[BD\times CD\]
C) \[BC\times AC\]
D) \[AB\times BC\]
Correct Answer: B
Solution :
In \[\Delta 's\]CDA and BAD. we have \[\angle BAD={{90}^{o}}-\angle ABD\] \[\angle DAC={{90}^{o}}-\angle BAD\] \[={{90}^{o}}-({{90}^{o}}-\angle ABD)=\angle ABD\] \[\Rightarrow \] \[\angle DAC=\angle ABD\] Now, in \[\Delta 's\] BAD and ADC, we have \[\angle ABD=\angle CAD\] \[\angle BDA=\angle ADC\] [Each \[{{90}^{o}}\]] \[\therefore \] \[\Delta BDA\tilde{\ }\Delta ADC\] [By AA similarity] \[\Rightarrow \] \[\frac{BD}{AD}=\frac{AD}{DC}\] \[\Rightarrow \] \[A{{D}^{2}}=BD\times CD\]You need to login to perform this action.
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