Congruent triangles
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com O^a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-67e5e14447698c4cf3cf6e13bc91c292_l3.png)
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com OO^a=b](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-131755b7585263d6315f468680310ae2_l3.png)
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.
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com O^b](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-52d34ae3af0de9b9790142dd86266a7b_l3.png)
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
![Rendered by QuickLaTeX.com OO^b=a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-03247f6f0e7d9d47b1594d6e21e7566c_l3.png)
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).
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com O^a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-67e5e14447698c4cf3cf6e13bc91c292_l3.png)
![Rendered by QuickLaTeX.com O^b](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-52d34ae3af0de9b9790142dd86266a7b_l3.png)
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com CA](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-250879d8959d81e68a48d0fcd4f28fdf_l3.png)
![Rendered by QuickLaTeX.com \triangle OO^aO^b](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-06b0a0a3d00030713b794c22cd31a533_l3.png)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
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
![Rendered by QuickLaTeX.com O^a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-67e5e14447698c4cf3cf6e13bc91c292_l3.png)
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com AOCO^a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-76e028b37bb85b9a05a74b0c31b1ec7b_l3.png)
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
.
![Rendered by QuickLaTeX.com O^a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-67e5e14447698c4cf3cf6e13bc91c292_l3.png)
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com BC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c4f3ce61859dcd00e52fbf771b86cb55_l3.png)
![Rendered by QuickLaTeX.com O^c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-1425d5b43f3ede9df61dbf5f7a314569_l3.png)
![Rendered by QuickLaTeX.com O](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-cc7fd4cac66bf4144e6ccd78d2ae8570_l3.png)
![Rendered by QuickLaTeX.com AB](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b7694a0a3a6ff6caddf255cb508f628_l3.png)
![Rendered by QuickLaTeX.com AO^cHO^a](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-07c34660ba1fece468a20db3621eca7a_l3.png)
![Rendered by QuickLaTeX.com H](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d2121613062e686e294d3867f5da2955_l3.png)
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.
Task
- (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:
is congruent to
is isosceles with
is isosceles with
is right angled at
is the circumcenter of
is right-angled at
is right-angled at
- quadrilateral
is a rectangle
- the points
are concyclic with
as diameter
- the reflection
of
over
lies internally on
- the reflection
of
over
lies externally on
- radius
is parallel to side
is the reflection of
over side
- the nine-point center lies on
- the orthic triangle is isosceles with
- the geometric mean theorem holds
- the bisector of
has length
, where
- the orthocenter is a reflection of vertex
over side
- segment
is tangent to the circumcircle at point
- median
has the same length as the segment
- the bisector
of
is tangent to the nine-point circle at
is a convex kite with diagonals
and
- altitude
is tangent to the nine-point circle at
- 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
.