*same*vertex. Then compute their product. If the product equals the area of the triangle, then the triangle

*must*be a right triangle.

The above property is exclusive to right triangles, and there are many other characterizations of right triangles, as seen in this list.

Today’s post will push at least two more entries into the list, in addition to that of July 14 .

Consider a right triangle in which as shown below:

The length of the median to the hypotenuse is half the length of the hypotenuse: . Thus, the area is:

Suppose that the area of is equal to the product of the altitude and median from the same vertex . Let be the side-lengths and let be the circumradius. Then

The area can be given as , and itself can be written as , by the extended law of sines. So:

Let be the circumradius of the parent triangle . If the circle with diameter coincides with the nine-point circle, then we must have , since the radius of the nine-point circle is .

the latter equation is one of the characterizations of right triangles given here.

In the next example we have a weaker requirement — we just ask that the nine-point circle pass through or .

Suppose that the nine-point circle passes through . Let be the nine-point center. Then is a radius of the nine-point circle, namely . But then .

Similarly, if the nine-point circle of a triangle passes through the orthocenter of the triangle, then the parent triangle must be a right triangle.

of is

*half*the square of the harmonic mean of and . PROVE that .

The *harmonic mean* of and is . The square of the length of the bisector of is given by . We get

and then finally after simplifications.

## Takeaway

In , let be the side-lengths, the median from vertex , the altitude from vertex , the orthocenter, and the circumcenter. Then the following statements are *equivalent*:

- the area of equals
- is a
*right*triangle with - the nine-point circle of passes through
- the nine-point circle of passes through
- the circle with diameter coincides with the nine-point circle of
- is the geometric mean of the two equal segments it creates on the opposite side
- the square of the length of the bisector of angle is half the square of the harmonic mean of and .

In the case of statement *6*, one can replace the geometric mean with arithmetic mean or harmonic mean. Why?

## Task

- (Early 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 , and the reflection of over side . PROVE that the following*fifty-four*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 .