question_answer

A point D is taken from the side BC of a right angled \[\Delta ABC,\]where AB is hypotenuse. Then,

A) \[A{{B}^{2}}+C{{D}^{2}}=B{{C}^{2}}+A{{D}^{2}}\]

B) \[C{{D}^{2}}+B{{D}^{2}}=2\,A{{D}^{2}}\]

C) \[A{{B}^{2}}+A{{C}^{2}}=2\,A{{D}^{2}}\]

D) \[A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}\]

Correct Answer: A

Solution :

[a] In \[\Delta ABC,\] \[A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}\]               (i) In \[\Delta ACD,\] \[A{{D}^{2}}=A{{C}^{2}}+C{{D}^{2}}\] \[\Rightarrow \]   \[A{{C}^{2}}=A{{D}^{2}}-C{{D}^{2}}\]                (ii) From Eqs. (i) and (ii), we get \[A{{B}^{2}}=A{{D}^{2}}-C{{D}^{2}}+B{{C}^{2}}\] \[\Rightarrow \]   \[A{{B}^{2}}+C{{D}^{2}}=A{{D}^{2}}=A{{D}^{2}}+B{{C}^{2}}\]