In a quadrilateral $A B C D,$ if the diagonals $A C$ and $B D$ intersect at right angles, then:

A B^{2} + B C^{2} = D C^{2} + D A^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A B^{2} + C D^{2} = B C^{2} + D A^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A B^{2} + A D^{2} = C B^{2} + C D^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $A B^{2} + C D^{2} = B C^{2} + D A^{2}$

Consider $△ B O C$
$\Rightarrow O B^{2} + O C^{2} = B C^{2} ⟶ (1)$
Similarly from $△ A O B,$ $△ A O D$ $ξ$ $△ C O D$
$\Rightarrow O B^{2} + O A^{2} = A B^{2}$
$O A^{2} + O D^{2} = A D^{2}$
$O C^{2} + O D^{2} = C D^{2}$
Adding all the equations,
$\Rightarrow 2 (O B^{2} + O C^{2} + O D^{2} + O A^{2}) = B C^{2} + A B^{2} + A D^{2} + C D^{2}$
$2 B C^{2} + 2 A D^{2} = B C^{2} + A B^{2} + A D^{2} + C D^{2}$
$\Rightarrow A B^{2} + C D^{2} = B C^{2} + D A^{2}$
Hence, the answer is $A B^{2} + C D^{2} = B C^{2} + D A^{2} .$

Suggest Corrections

Similar questions

A B^{2} + C D^{2} = B C^{2} + D A^{2}

A B C D, A B^{2} + B C^{2} + C D^{2} + D A^{2} =

A C

B D

O

A B^{2} + B C^{2} + C D^{2} + D A^{2} = 4 (O A^{2} + O B^{2})

{AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} + 4 {PQ}^{2}

A B C D, ∠ B = 90^{o}

∠ D = 90^{o}

2 {A C}^{2} - {A B}^{2} = {B C}^{2} + {C D}^{2} + {D A}^{2}

Join BYJU'S Learning Program