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# An inequality in cyclic quadrilaterals

In a cyclic quadrilateral with diagonals and and circumradius , the quantity is an upper bound for the sum of the squares of the diagonals; in other words:

It is only when the parent quadrilateral is a rectangle that equality holds.

## Easy proofs

Let be a cyclic quadrilateral with circumradius and diagonals , . PROVE that .

Since triangles and have the same circumradius we have from the above diagram that and . Thus .

Let be a cyclic quadrilateral with circumradius and diagonals , . PROVE that .

Follows from the previous example, since the opposite angles at and , as well as at and , ensures that and .

Let be a cyclic quadrilateral with circumradius and diagonals , . PROVE that .

We have , . Thus .

Let be a cyclic quadrilateral with circumradius and diagonals , . Deduce that .

Since from the preceding example, and , it follows that .

## Exclusive property

For rectangles only.

Let be a cyclic quadrilateral with circumradius and diagonals , . If , PROVE that is a rectangle.

We always have . If the right side is now , then we must have . The latter equation can only be satisfied by taking and . Similarly we have , and so . We take . Thus, in such a cyclic quadrilateral we have .

Jump straight to the exercises at this stage — or read what follows for a little bit of distraction. As we promised previously, we’ll be bringing you occasional updates about our ongoing hide and seek — or, tug-of-war — game with some unknown bots. The game is exciting because we currently have the ascendancy.

## Takeaway

Let be a convex cyclic quadrilateral with circumradius and diagonals , . Then the following statements are equivalent:

1. is a rectangle
2. .