Question

Diagonals of rhombus ABCD intersect each other at point O. Prove that :
$O{A}^2+O{C}^2=2A{D}^2-\frac{B{D}^2}{2}$
Answer: $O{A}^2+O{C}^2=2A{D}^2-\frac{B{D}^2}{2}$
If true then enter 1 else if False enter 0

Answer 1

Given: ABCD is a rhombus. Diagonals meet at O.
Diagonals of a rhombus bisect each other at right angles.
Hence, in $△OAB$
$O{A}^2+O{B}^2=A{B}^2$..(I)
In $△OBC$.
$O{B}^2+O{C}^2=B{C}^2$..(II)
Adding I and II,
$O{A}^2+O{C}^2=A{B}^2+B{C}^2-2O{B}^2$
$O{A}^2+O{C}^2=2A{D}^2-2\left(\frac{B{D}^2}{4}\right)$ (Since, ABCD is a rhombus, sides are equal and diagonal bisect each other)
$O{A}^2+O{C}^2=2A{D}^2-\frac{B{D}^2}{2}$