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Approximate Pythagorean Identity II

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)

In a non-right , let be the foot of the altitude from vertex , and let (or ). If the side-lengths satisfy equation (1), PROVE that .

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:

Consider with vertices at , , and . If is the foot of the altitude from vertex and , verify that .

From the diagram below

we get . Equation (1) holds:

as well as equation (2), because:

Consider with vertices at , , and . If is the foot of the altitude from vertex and , verify that .

Note that and can switch places. From the diagram below

we get . Equation (1) holds:

as well as equation (2), because:

Consider with vertices at , , and . If is the foot of the altitude from vertex and , verify that .

From the diagram below

we get . Notice that equation (1) holds:

as well as equation (2), since:

Consider with vertices at , , and . If is the foot of the altitude from vertex and , verify that .

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:

1. is isosceles with
2. is isosceles with
3. is the circumcenter of
4. 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:
1. is congruent to
2. is isosceles with
3. is isosceles with
4. is right angled at
5. is the circumcenter of
6. radius is parallel to side
7. the nine-point center lies on
8. the orthic triangle is isosceles with
9. the geometric mean theorem holds
10. the bisector of has length , where
11. the orthocenter is a reflection of vertex over side
12. segment is tangent to the circumcircle at point
13. the bisector of is tangent to the nine-point circle at
14. is a convex kite with diagonals and
15. altitude is tangent to the nine-point circle at
16. 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 .