![Rendered by QuickLaTeX.com 120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-db28b4ad2dd28af7ee51c281db14a9f5_l3.png)
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.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com \angle A=\angle C=30^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-408e55b18738970715ba119fa33d44b9_l3.png)
![Rendered by QuickLaTeX.com \angle B=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-09aa04cb2daf52ea1c2a9cb2b19f19f8_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com AB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b7694a0a3a6ff6caddf255cb508f628_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
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
.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com AB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b7694a0a3a6ff6caddf255cb508f628_l3.png)
![Rendered by QuickLaTeX.com A](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3404ae68a19d10407c93ea3824613e4e_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com \angle A=\angle C=30^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-408e55b18738970715ba119fa33d44b9_l3.png)
![Rendered by QuickLaTeX.com \angle B=120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-09aa04cb2daf52ea1c2a9cb2b19f19f8_l3.png)
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
.
![Rendered by QuickLaTeX.com a,b,c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-57e6624e316682b226e631d7234f001d_l3.png)
![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
![Rendered by QuickLaTeX.com K](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-45553b59c9922b4b50444b81c7a16595_l3.png)
![Rendered by QuickLaTeX.com K=\frac{1}{2}ac\cos A=\frac{1}{2}ac\cos C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-8df9467695efa29d3dbf84052941a6dc_l3.png)
![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
![Rendered by QuickLaTeX.com 30^{\circ}-30^{\circ}-120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0a10497c2335a5cd89bea71b5766548a_l3.png)
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, .
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
![Rendered by QuickLaTeX.com OABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-838116fdb1d6c6685fa9f3ba70c10aa3_l3.png)
![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
![Rendered by QuickLaTeX.com 30^{\circ}-30^{\circ}-120^{\circ}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0a10497c2335a5cd89bea71b5766548a_l3.png)
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.
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b5002f05fd90c80f81a9e0e9c845e02d_l3.png)
![Rendered by QuickLaTeX.com OABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-838116fdb1d6c6685fa9f3ba70c10aa3_l3.png)
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:
- the reflection of
over
is
- the area
satisfies
- the reflection of
over
is
and the reflection of
over
is
Task
- (Aufbau) In triangle
, let
be the side-lengths,
the circumradius,
the circumcenter,
the nine-point center, and
the orthocenter. PROVE that the following statements are equivalent:
- the reflection of
over
is
- the area
satisfies
- the circle with diameter
passes through vertices
and
- radius
is parallel to side
and radius
is parallel to side
- the reflection of
over
is
and the reflection of
over
is
is the geometric mean of
and
, and
is the geometric mean of
and
. (
and
are the feet of the altitudes from
and
, respectively.)