question_answer

In an isosceles \[\Delta \,ABC,\] \[AB=AC\] and \[BD\bot AC\] then \[(B{{D}^{2}}-C{{D}^{2}})=\]

A)  \[2CD\text{ }.\text{ }AD\]                 

B)  \[2AC.BD\]    

C)  \[2CD.CD\]    

D)         \[2AB.BC\]

Correct Answer: A

Solution :

Given : \[\Delta ABC\]in which AB = AC and \[BD\bot AC.\] From right \[\Delta ADB,\]we have             \[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}\]             [By Pythagoras theorem] \[\Rightarrow \] \[A{{C}^{2}}=A{{D}^{2}}+B{{D}^{2}}\]   \[[\because \,\,AB=AC]\] \[\Rightarrow \] \[{{(CD+AD)}^{2}}=A{{D}^{2}}+B{{D}^{2}}\]                             \[[\because \,\,AC=CD+AD]\] \[\Rightarrow \] \[C{{D}^{2}}+A{{D}^{2}}+2CD.AD=A{{D}^{2}}+B{{D}^{2}}\]             \[\Rightarrow \] \[(B{{D}^{2}}-C{{D}^{2}})=2CD.AD\]