During one of my recent mathematical explorations, I wanted to calculate the distance of the points of an isosceles triangle to its orthocentre (points where the altitudes meet).
In other words, I firstly wanted to find in terms of in the following figure.
Here is the approach I initially came up with. Since the angles at and are each degrees, quadrilateral is cyclic. Then by the intersecting chords theorem (or by the similarity of triangles and ),
from which …(1)
Also, since triangles and are similar,
from which …(2)
Combining (1) and (2),
It follows that . Who would have thought that such a simple answer would be found! Hence, we have rediscovered a way of constructing the reciprocal of a positive number : draw an isosceles triangle with base length and height , then the height of its orthocentre above the base is .
Naturally, the question comes up of whether there is an easier way of calculating this answer and surely enough there is. If we extend until it meets the circumcircle of at , we find that are equal (each being the complement of ). Then we also have , both angles lying on the arc . Both acute and obtuse angled cases are illustrated below.
It follows by Angle-Angle-Side that triangles HPB and C’PB are congruent, from which . In other words, the reflection of in lies on the circumcircle! This fact is true for the reflection of in any side, and need not be isosceles.
By the intersecting chords theorem for and intersecting at (or equivalently, by the similarity of triangles and ),
or as required.
I later found that the second proof is also given in pp18-19 of Honsberger’s book entitled “Episodes in 19th and 20th century Euclidean Geometry”.