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# Nine-point center equals a vertex III

Earlier we saw that the nine-point center of a triangle coincides with one of the vertices of the triangle precisely when the parent triangle is isosceles with an apex angle of .

Equivalently, two radii are parallel to two sides of the parent triangle, as will be shown in examples 1 and 2 below.

Everything we’ve been considering so far seems to work smoothly for the -triangle.

In , let and . PROVE that the radius through is parallel to side and the radius through is parallel to side .

Since the parent triangle is obtuse, its circumcenter lies outside, opposite the obtuse angle. Thus is a (convex) quadrilateral.

If is the radius of the circumcircle of triangle , then we have that and (see here). Since is the circumcenter, we also have that . Hence, is a rhombus; in fact, a special rhombus with the property that the diagonal is equal to the equal sides. Every rhombus is a parallelogram so we conclude that radius is parallel to side and radius is parallel to side .

In , suppose that the radius through is parallel to side , and the radius through is parallel to side . PROVE that and .

Suppose that a triangle with circumcenter has the property that radius is parallel to side and radius is parallel to side . Such a triangle is necessarily obtuse, so that the circumcenter lies outside the triangle and we again have a quadrilateral . Join and let . Since is parallel to , we have that . Since is the circumcenter, triangle is isosceles and so . The situation is shown below, with the circumcircle included:

Since is parallel to , we have that . Thus, triangle is isosceles; moreover, . By the inscribed angle theorem:

Thus, the interior angles of triangle are and .

Let be the side-lengths of triangle , whose area satisfies . PROVE that triangle is the isosceles triangle.

Since the area satisfies , we have that .

Normally, . And so we must have . In turn, this implies

is not possible due to the fact that and . So we take , which yields . Using we get

Consequently, .

Let be the circumcenter of triangle . If the convex quadrilateral is a rhombus, PROVE that triangle is the isosceles triangle.

No matter how the vertices are arranged, we’ll end up with a rhombus in which a diagonal is equal to the side-lengths. The interior angles of such a rhombus are . Now just consider the triangle that remains when is removed.

Let be the circumcenter of triangle . If the convex quadrilateral is a rhombus, PROVE that the rhombus cannot be a square.

Consider various arrangements of the rhombus , for example the one shown below:

Apply the inscribed angle theorem to see that it is not possible to have be a square.

## Takeaway

In any triangle , let be the side-lengths, the orthocenter, the circumcenter, and the circumradius. Then the following statements are equivalent:

1. the reflection of over is
2. the area satisfies
3. the reflection of over is and the reflection of over is