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

## Easy proofs

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

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

We have , . Thus .

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

## Exclusive property

For rectangles only.

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 .

## Elusive password

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.

### Scene 1: Barrage by bots

### Scene 2: Breathe for a bit

### Is any of the above IP addresses familiar?

## Takeaway

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

- is a rectangle
- .

## Task

- (Special case) Let be a convex cyclic quadrilateral with vertices at , , , and . PROVE that:
- diagonal bisects diagonal
- , where and
- and , where is the length of the Newton line.