Question

State true or false:

In quadrilateral ABCD, the diagonals AC and BD intersect each each other at right angle, then $A{B}^2+C{D}^2=A{C}^2+D{A}^2$

• A. True
• B. False

Answer 1

The correct option is B False

Given: Quadrilateral ABCD, diagonals intersect at right angles at O
In $△OAB$,
$O{A}^2+O{B}^2=A{B}^2$ (I) (Pythagoras theorem)
In $△OBC$,
$O{B}^2+O{C}^2=B{C}^2$ (II) (Pythagoras theorem)
In $△OCD$,
$O{C}^2+O{D}^2=C{D}^2$ (III) (Pythagoras theorem)
In $△OAD$,
$O{A}^2+O{D}^2=D{A}^2$ (IV) (Pythagoras theorem)
Add I and III,
$O{A}^2+O{B}^2+O{C}^2+O{D}^2=A{B}^2+C{D}^2$ (V)
Add II and IV,
$O{A}^2+O{B}^2+O{C}^2+O{D}^2=B{C}^2+D{A}^2$ (VI)
Equating V and VI,
$A{B}^2+C{D}^2=B{C}^2+D{A}^2$