question_answer

In the given \[\Delta ABC,\] \[AD\bot BC\]and \[\angle A\] is right angled. Then \[A{{D}^{2}}=\]

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]             \[\Rightarrow \]  \[\frac{BD}{AD}=\frac{AD}{DC}\]  \[\Rightarrow \] \[A{{D}^{2}}=BD\times CD\]