It is only when the parent quadrilateral is a rectangle that equality holds.
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 .
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 .
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 is familiar?
- (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.