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# Reflecting the circumcenter I

In , let be the usual side-lengths and let be the circumcenter. Let denote the reflections of over sides respectively. Then is congruent to the parent . If satisfy the identity

(1)

then in addition we have that is congruent to .

## Congruent triangles

In a non-right , let be the reflection of the circumcenter over side . Then if and only if equation (1) holds.

Let’s prove one direction: (1). Consider shown below:

We showed previously that in any triangle with orthocenter . Thus . By one of the equivalent statements here, we conclude that equation (1) holds.

In a non-right , let be the reflection of the circumcenter over side . Then if and only if equation (1) holds.

Similar to example 1 above. If is the reflection of the circumcenter over side , then we always have in any triangle. This would then lead to , which by the equivalent statements here implies equation (1).

Let be the circumcenter of , and let , be the reflection of over side and side , respectively. Then is congruent to the parent , if equation (1) holds.

This follows from the two preceding examples, together with the fact that the segment has length . (Recall that is congruent to the parent in such a way that , , and .)

## Cyclic trapezium

Let be the reflection of the circumcenter over side . If equation (1) holds, then the quadrilateral is an isosceles trapezium, hence cyclic.

Note that under the assumption of equation (1), we have that the reflection lies on side . Let be the circumradius of the parent triangle . Then and also . Since radius is parallel to side (see here), we get an isosceles trapezium .

Let be the reflection of the circumcenter over side , and let be the reflection of over side . If equation (1) holds, then the quadrilateral is an isosceles trapezium, hence cyclic. (Here, is the orthocenter.)

Similar to example 4 above.

## Takeaway

In , let be the side-lengths, the circumcenter, the reflection of over side , the reflection of over side , and the reflection of over side . Then the following statements are equivalent:

As usual, we have more of these in the exercises.

• (Mid fifties) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the circumradius, the circumcenter, the nine-point center, the orthocenter, the midpoint of side , the reflection of over side , the reflection of over side , and the reflection of over side . PROVE that the following fifty-six statements are equivalent:
1. is congruent to
2. is isosceles with
3. is isosceles with
4. is right angled at
5. is the circumcenter of
6. is right-angled at
7. is right-angled at
9. the points are concyclic with as diameter
10. the reflection of over lies internally on
11. the reflection of over lies externally on
12. radius is parallel to side
13. is the reflection of over side
14. the nine-point center lies on
15. the orthic triangle is isosceles with
16. the geometric mean theorem holds
17. the bisector of has length , where
18. the orthocenter is a reflection of vertex over side
19. segment is tangent to the circumcircle at point
20. median has the same length as the segment
21. the bisector of is tangent to the nine-point circle at
22. is a convex kite with diagonals and
23. altitude is tangent to the nine-point circle at
24. segment is tangent to the nine-point circle at .
( short of the target.)
• (Extra feature) If satisfies equation (??), PROVE that its nine-point center divides in the ratio .