If a non-right triangle, like the one shown above, satisfies the identity
(1)
then we can recast equation (1) in a form that looks more like the usual Pythagorean identity, namely:
(2)
As , we have an approximate Pythagorean identity
(3)
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com s=BF_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d6590ae0826cd33c08391b68a68ab07f_l3.png)
![Rendered by QuickLaTeX.com s=AF_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a8999d2b41845ca97178dd0f76449db6_l3.png)
![Rendered by QuickLaTeX.com a,b,c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-57e6624e316682b226e631d7234f001d_l3.png)
![Rendered by QuickLaTeX.com (c+2s)^2=a^2+b^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a6fbea5c461a02fdcdc341a88fec6899_l3.png)
Consider the following diagram:
Let be the circumradius and suppose that
(if
, take
). We’ll use some properties that are consequences of equation (1), including the one from our previous post:
Then:
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a4c057a4f58741a4e3fc4d5121ae3540_l3.png)
![Rendered by QuickLaTeX.com B(3,6)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b67631b2dbdedc7cc3d349be4e9a2d8_l3.png)
![Rendered by QuickLaTeX.com C(0,10)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-f5c0cc9017e2a098d03651147c899333_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com s=BF_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d6590ae0826cd33c08391b68a68ab07f_l3.png)
![Rendered by QuickLaTeX.com \left(c+2s\right)^2=a^2+b^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4616263c29d4e7d8f1dbd623d2ca5467_l3.png)
From the diagram below
we get . Equation (1) holds:
as well as equation (2), because:
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a4c057a4f58741a4e3fc4d5121ae3540_l3.png)
![Rendered by QuickLaTeX.com B(3,-9)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-43212633f3c663edcc119dfc80752010_l3.png)
![Rendered by QuickLaTeX.com C(5,5)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-5bae280ade7d95e5b5374fdea52bf0a7_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com s=AF_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a8999d2b41845ca97178dd0f76449db6_l3.png)
![Rendered by QuickLaTeX.com \left(c+2s\right)^2=a^2+b^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4616263c29d4e7d8f1dbd623d2ca5467_l3.png)
Note that and
can switch places. From the diagram below
we get . Equation (1) holds:
as well as equation (2), because:
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(-6,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-79c16eb208dc820695dfd9a30fa1c847_l3.png)
![Rendered by QuickLaTeX.com B(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-7045877de833cb7e53ba1ef4ddfaeb24_l3.png)
![Rendered by QuickLaTeX.com C(2,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-e2fb9eee47c72cdf6ab23e648c29df8c_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com s=BF_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d6590ae0826cd33c08391b68a68ab07f_l3.png)
![Rendered by QuickLaTeX.com (c+2s)^2=a^2+b^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-a6fbea5c461a02fdcdc341a88fec6899_l3.png)
From the diagram below
we get . Notice that equation (1) holds:
as well as equation (2), since:
![Rendered by QuickLaTeX.com \triangle ABC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-b8aadc08f11d62c883914b8be4436356_l3.png)
![Rendered by QuickLaTeX.com A(4,4)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4acc2a75b5891b8bc98456dcbc05edc4_l3.png)
![Rendered by QuickLaTeX.com B(0,0)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-7045877de833cb7e53ba1ef4ddfaeb24_l3.png)
![Rendered by QuickLaTeX.com C(1,-2)](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-c0508a321bf286d7f4b8683a94a1694c_l3.png)
![Rendered by QuickLaTeX.com F_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-540652df9e9a6c584f9d7c1a034ecc35_l3.png)
![Rendered by QuickLaTeX.com C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3238eb6669792857eccb924606e71d82_l3.png)
![Rendered by QuickLaTeX.com s=BF_C](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-d6590ae0826cd33c08391b68a68ab07f_l3.png)
![Rendered by QuickLaTeX.com \left(c+2s\right)^2=a^2+b^2](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-4616263c29d4e7d8f1dbd623d2ca5467_l3.png)
From the diagram below
we get . Equation (1) holds:
as well as equation (2), since
As in this post, notice that the title of today’s post can be abbreviated as API, an absolutely endearing term in computer programming.
Takeaway
In a non-right , let
be the side-lengths,
the circumradius,
the circumcenter,
the orthocenter, and
the nine-point center. Then the following statements are equivalent:
is isosceles with
is isosceles with
is the circumcenter of
is right angled at
.
Don’t confuse with the functional group of an aldehyde.
Task
- (Early fourties) 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, and
the orthocenter. PROVE that the following forty four statements are equivalent:
is congruent to
is isosceles with
is isosceles with
is right angled at
is the circumcenter of
- radius
is parallel to 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
- 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
.
(Wow!!! The numbers increased so rapidly, from twenty one a few days ago.)
- (Extra feature) If
satisfies equation (1), PROVE that its nine-point center
divides
in the ratio
.