The diagonals AC and BC of a quadrilateral ABCD intersect at O, at right angles. Prove that AB²+CD²=AD²+BC².

Open in App

Solution

g

Suggest Corrections

Similar questions

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

A B C D

E . A E = 10, B E = 8, C E = 6,

D E = 4

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

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

A B C D

∠ B = 90

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

∠ A C D = 90^{0}

In a quadrilateral ABCD, $∠ B = 90^{\circ}, A D^{2} = A B^{2} + B C^{2} + C D^{2}$ , prove that $∠ A C D = 90^{\circ}$

A B C D

∠ A B C = 90^{o}

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

∠ A C D = 90^{o}

Join BYJU'S Learning Program