question_answer

In a triangle ABC, if angle B = \[90{}^\circ \] and D is the point in BC such that BD = 2 DC, then

A) \[A{{C}^{2}}=\,A{{D}^{2}}+3C{{D}^{2}}\]

B) \[A{{C}^{2}}=\,A{{D}^{2}}+5C{{D}^{2}}\]

C) \[A{{C}^{2}}=\,A{{D}^{2}}+7C{{D}^{2}}\]

D) \[A{{C}^{2}}=\,A{{B}^{2}}+5B{{D}^{2}}\]

Correct Answer: B

Solution :

 From right angled \[\Delta \,\,ABC,\]                 \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]                      ??.(i) Also, from right angled \[\Delta \,ABD,\]                 \[A{{D}^{2}}=A{{B}^{2}}+B{{D}^{2}}\] or            \[A{{B}^{2}}=A{{D}^{2}}-B{{D}^{2}}\] Substituting the value in (i), we get                 \[A{{C}^{2}}=(A{{D}^{2}}-B{{D}^{2}})+B{{C}^{2}}\]                 \[=A{{D}^{2}}+(B{{C}^{2}}-B{{D}^{2}})\]                 \[=A{{D}^{2}}+(BC-BD)\,(BC+BD)\] or            \[A{{C}^{2}}=A{{D}^{2}}+CD(3\,CD+2\,CD)\] or            \[A{{C}^{2}}=A{{D}^{2}}+5\,C{{D}^{2}}\]