Let be the nine-point center of triangle . We showed in our previous post that if the parent triangle is right-angled with , then
We’ll now prove what holds more generally for any triangle:
being the circumradius and the usual side-lengths.
Let’s first be reminded of the main ingredient: Stewart’s theorem. It states that given a triangle with side-lengths and a cevian like the one depicted below:
then we have:
In with the usual notation, PROVE that the orthocenter-circumcenter distance satisfies .
It was from this amazing site that we first learnt of the above identity. We’ll add our own proof of it to an already existing pool of proofs, using Stewart’s theorem as a tool. We’ll also use the formula for median lengths, and the fact that the centroid divides a median in the ratio (measured from a vertex) and also divides the Euler line in the ratio (measured from the orthocenter).
Consider the diagram below where is the centriod of , is the midpoint of , and (circumradius):
Notice our new point just andering along side . More importantly, observe that is perpendicular to ( is the midpoint of and is the circumcenter). Therefore, the Pythagorean theorem applied to gives:
The above procedure works regardless of the vertex we started with. And it also works even in the peculiar case where are co-linear. It also doesn’t matter if both and are outside the parent triangle.
Let be the orthocenter of with circum-radius . PROVE that .
Apply Stewart’s theorem to in the diagram below:
Add all the three last equations:
Let and be the nine-point center and orthocenter of . PROVE that .
We only slightly modify the proof given in example 2, replacing the centroid with the nine-point center (note that is the midpoint of ).
In above, is a median, so:
Similarly, if we connect the segment to vertices and instead, we obtain (with and as medians)
Add the expressions for and simplify:
Since , we have . In turn:
PROVE that , where is the nine-point center of . Hence deduce that in a right triangle with hypotenuse of length , the sum of the squares of the distances from the vertices to the nine-point center is the length of the hypotenuse.
We obtained as an intermediate step in example 3 above.
Now if is such that is the hypotenuse with length , then . The circumradius is half of the length of the hypotenuse, so . Substituting in the expression for , we obtain:
We show in example 5 below that the above relation also holds in a non-right triangle setting.
Consider in which and . PROVE that .
(If you’re curious as to how we obtained , see example 7 for a sample procedure.)
A key ingredient in the proof is the fact that the circumradius can be expressed as as per the extended law of sines.