math problem

[Debasish]

In a quadrilateral ABCD, with unequal sides if the diagonals AC and BD intersect at right angles, then
AB²+BC²=CD²+DA²
AB²+CD²=BC²+DA²
AB²+AD²=BC²+CD²
AB²+BC²=2(CD²+DA²)

 

[JPJ]

Strategy:

- Draw a figure, such as the attachment;

- Equations to be written out of the 4 pythagoric triangles in the figure:

(I) AB² = x² + w²
(II) BC² = x² + y²
(III) CD² = y² + z²
(IV) AD² = w² + z²

- By checking each alternative, one quickly sees that the second one (AB²+AD²= BC²+CD²) is the only possible, since (I) + (III) = (II) + (IV).

 

[arunroy15]

answer would be (iii)

strategy:

- draw a figure, such as the attachment;

- equations to be written out of the 4 pythagoric triangles in the figure:

(i) ab² = x² + w²
(ii) bc² = x² + y²
(iii) cd² = y² + z²
(iv) ad² = w² + z²

- by checking each alternative, one quickly sees that the second one (ab²+ad²= bc²+cd²) is the only possible, since (i) + (iii) = (ii) + (iv).

ab^2+cd^2= bc^2+cd^2