question_answer

In a quadrilateral ABCD, if the diagonals AC,BD intersect at right angles, then

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

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

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

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

Correct Answer: B

Solution :

 From right angled triangles \[B{{C}^{2}}=O{{C}^{2}}+O{{B}^{2}}\]     ?..(i) \[C{{D}^{2}}=O{{D}^{2}}+O{{C}^{2}}\]    ?..(ii) \[A{{D}^{2}}=A{{O}^{2}}+O{{D}^{2}}\]   ??(iii) \[A{{B}^{2}}=A{{O}^{2}}+O{{B}^{2}}\]         ?..(iv) From equations (ii) and (iv), \[A{{B}^{2}}+C{{D}^{2}}=A{{O}^{2}}+O{{D}^{2}}+O{{C}^{2}}+O{{B}^{2}}\]                 \[=A{{D}^{2}}+B{{C}^{2}}\]