Isosceles triangles

A triangle with two sides equal is described as isosceles. Since each side of a triangle meets each of the other two sides in a corner, the two equal sides meet; if we bisect the angle between these sides and reflect the triangle in the bisector line, we simply interchange the two equal sides; in particular, we interchange their end-points, so the mirror image of the third side is itself. For now, I won't fully draw that third edge, because it'd too obviously have a property I haven't yet proved it'll have. However, the angles (marked with blue arcs) it forms with the first two edges must in fact be equal, since each is the image of the other in a reflection.

We have reflected the third edge in a line that cuts it, though: we know the mirror line bisects the angle between the first two edges, so the third edge's two end-points are on opposite sides of the mirror, obliging the edge to cut the mirror in order to get from one to the other. When a line crosses a mirror, on a given side of the crossing line, their meeting makes two angles (marked in blue) that add up to a half turn. The crossing line's mirror image, on its matching side, makes the same two angles (in orange) but reverses which side of the mirror it makes them on. For the image of the line to coincide with the line, the two angles that add up to a half turn must thus be equal, making each a quarter turn, i.e. a right angle. Furthermore, since the two parts into which the mirror line cuts the third edge are each other's mirror images in the mirror line, they must be of equal length, making the bisector of the angle between two equal edges of a triangle the perpendicular bisector of the third edge.

Thus, in fact, that third edge of the isosceles triangle meets the angle bisector (mirror line) in a right angle. So, when a triangle has two sides equal, the angles in which they meet the third side must also be equal; and the bisector of the angle between the two equal sides meets the opposite edge in a right angle. For all that the result may be obvious, pause to consider a triangle in which we're given that two internal angles are equal: if we drop a perpendicular from the third corner onto the line joining the two equal angles, cutting the triangle into two right angle triangles, each with one of the two equal angles in it; so their respective angles at the third corner (being the rest of a half turn, aside from the right angle and one of the equal angles) must be equal; and these are the angles the perpendicular makes with the sides from the third corner to the two equal angles; so, in fact, the perpendicular bisects the angle at the third corner. Since the perpendicular meets the edge between the two equal angles in a right angle, reflection in the perpendicular maps the line, of which this edge is a part, onto itself; since the perpendicular bisects the angle opposite this edge, it maps the lines of which the other two edges are segments onto one another; since the corners with the equal angles are determined by where their joining line meets the other two edges, this ensures the reflection interchanges the two equal corners, so the two sides from the third corner to each of these are equal. Thus a triangle with equal internal angles connects these two angles to its third corner with equal sides, making it isosceles. So having two angles equal implies and is implied by having two sides equal.

A triangle with no two sides equal is described as scalene.

Inside a circle

Now consider a chord of a circle; this is a line that runs from one point on the circle to another. Naturally, we can connect each end of it to the centre with straight lines; as its ends lie on the circle, these lines are radii of the circle, of equal length; so, with the chord, they form an isosceles triangle; and they meet the chord in equal angles. The radii meet at the centre in an angle that's half a turn minus the sum of the two equal angles in which the radii meet the chord; i.e. half a turn minus twice the angle between either radius and the chord.

Next, let's connect the two ends of our chord to a third point on the circle, forming a triangle with all three corners on the circle. The two edges we've just added are, of course, chords. The radius to the new vertex cuts the angle of our triangle at that new vertex into two parts, between the radius and each of the two new chords. The angles at the centre between this radius and the two radii to the ends of our original chord are thus half a turn minus twice the new chords' respective parts of the new vertex's angle; if we add these two angles together, at the centre, they come to (two halves make …) a whole turn minus twice the angle in the new vertex's corner; the angle between the original chord's two end-radii, where they meet at the centre, is the rest of a turn so must in fact be equal to twice the angle in the new corner.

The reasoning for that depended on a diagram in which the circle's centre was inside the triangle formed by the original chord and the new vertex. Let's just check what happens when the centre lies outside the triangle. First, look at the case where the new vertex is on the same side of the chord as the circle's centre. If the circle's centre is outside the resulting triangle, then the triangle's angle at the new vertex is the difference between, rather than sum of, the angles between the radius to the new vertex and the two chords that connect it to the ends of the original. In that case, the angle between the other two radii is likewise the difference between their respective angles to this new radius; each of these last is just a half turn minus twice an angle at the new vertex, so the difference is just twice the difference between the new vertex's angle between radius and edges, i.e. the angle at the centre, facing the original chord, is as before, just twice the internal angle of the triangle at the new vertex. The remaining case to consider has the new vertex on the far side of the chord from the centre of the circle. In this case, we can actually use the same reasoning as originally; it just leads to twice the new vertex's angle being the angle on the far side of the centre.

A A 2.A B B 2.B

In each case an angle at the centre is twice an angle at the new vertex; and these two angles lie on the same side of the centre and vertex, respectively. In particular, since the angle in the centre doesn't depend on where the new corner is, the angle in the new corner only depends on the chord opposite it, not on where around the circle we chose to put the new corner – as long as it's on the same side of the chord. One special case is particularly worthy of mention: when the angle at the centre is a half turn, so the radii towards the ends of the chord go in opposite directions, they necessarily lie along the chord – which is thus a diameter; any triangle formed of a diameter of a circle and a third point on the circle has a quarter turn, i.e. a right angle, at that third point.

This last is known as Thales' theorem, dating from around 600 BCE. As a special case, it has a separate and simpler proof, that considers all rectangles whose diagonal is the diameter of our circle. Each of these does have a right angle in each of its corners. The two diagonals of a rectangle are necessarily equal, being each other's mirror images in either of the edge bisectors of the rectangle; and these two equal diagonals meet at the centre of the rectangle, where they cross those edge-bisectors. Reflecting in first one then the other edge-bisector swaps the two halves of each diagonal, showing that their crossing point is their centre. Consequently, all the corners of the rectangle lie on the circle, not just the two that did so by consruction. There is therefore a right-angle triangle whose ends are the ends of our diagonal and whose third corner is the point we picked on the circumference.

A A B B One useful consequence of the angle subtended by a chord at the centre being twice the angle subtended at a point on the circle is that the perpendicular bisector of the chord, which is also the bisector of the angle it subtends at the centre of the circle, makes the same angle with a radius to its end as the chord subtends at a point on the (relevant side of the) circle. Each half of the chord is thus the side of a right-angle triangle opposite the given angle; and the right-angle triangle's hypotenuse is the circle's radius, so the chord's length is simply the diameter of the circle times the sine of the angle the chord subtends at a point on the circumference of the circle.

When the angle the chord subtends is more than a right angle, it's equal to the exterior angle of the right-angle triangle, opposite half the chord; but Sin(turn/2 −B) = Sin(B) for any angle B. Indeed, turn/2 −B is, in this case, the angle the chord subtends on the other side of the circle; and it naturally makes no difference which side of the chord we determine the length of. In the case where the subtended angle is a right angle, whose sine is 1, the chord is indeed a diameter.

This result is not infrequently useful in proving results about figures inscribed in a circle. It is also, equivalently, a proof of The Sine Rule.

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