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.

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 anglew 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.

A A B B One useful consequence of this 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.

Intersecting chords

Now consider two chords that cross each other. Each cuts the other into two pieces; what I'll be showing is that a rectangle whose sides are the two pieces of either chord has the same area as the corresponding rectangle made of the pieces of the other chord. (Representative rectangles are here shown in pink and orange, along with quarter-turn arcs about the point of intersection to show which side of each is equal to the other part of a chord; the two rectangles have equal areas). This is known as The Intersecting Chords Theorem.

B B A A p q r s Let's have a second copy of the diagram, without the rectangles, to decorate with construction lines and markings, to aid the proof; we can start with labels on the parts of the intersecting chords, indicating their lengths, splitting one as p+q and the other as r+s; the theorem claims p.q = r.s. If we traverse the circle noting the order in which we meet the ends of our chords, we alternate chords, since the two chords cross each other. We can draw a bounding quadrilateral that joins up the four chord-ends, in the same order as the circle visits them; the intersecting chords are diagonals of this quadrilateral. I've drawn that (in grey) with one pair of opposite sides dashed and the other pair solid. Each side of this quadrilateral is a chord of the circle, of course; so angles subtended by it at the other two corners of the quadrilateral must be equal; I've labelled the angles opposite dashed sides with letters; equal angles have the same letter.

This shows that the triangle, formed by either solid side and the two parts of the intersecting chords on its side of the intersection, has the same angles in two of its corners as the other solid side's matching triangle; and the third angle in each triangle is just the angle (the same on both sides) between the intersecting chords where they meet. Thus these two triangles are (mirrored) similar and the lenghts of their corresponding edges are in a common proportion; the two solid sides of our bounding quadrilateral correspond, for these purposes; edges from the intersection of the original chords correspond precisely if their other ends meet that solid side in the same angle. Thus each triangle's part of one intersecting chord corresponds to the other triangle's part of the other chord; the ratios of the corresponding parts being equal, p/r = s/q, then implies that the product of the two parts into which either intersecting chord is cut is the same as the product for the two parts of the other chord, p.q = r.s, which is exactly what the theorem claims.


How about the converse ? Whenever two line segments cross each other and the products of lengths of the pieces into which they cut each other are equal, we can construct the circumcircle of the triangle formed by any three of the end-points. Since we've left out only one of the four end-points, both end-points of one of the crossing lines lie on the circle, as does one end-point of the other; the crossing-point of the two lines is in the interior of the line segment that has both ends on the circle; so construct a line from the third point, that's on the circle by construction, through the crossing-point until it hits the circle again; this makes a second chord crossing the first, to which we can apply the intersecting chords theorem, revealing that the extension past the crossing point was exactly by the distance from the crossing-point to the fourth end-point of the original line segments, which is thus the other end of the chord, so lies on the circle. So, indeed, whenever two line segments cross, cutting each other into poieces whose products of lengths are equal, the circle through any three of the end-points must pass through the fourth as well.

When we apply the intersecting chords theorem to a chord and its perpendicular bisector chord, which is necessarily a diameter, the two halves of the bisected chord give us a square whose area is equal to that of a rectangle whose sides are the two parts into which it cuts the diameter. By constructing the diameter out of two sides of a rectangle, the intersecting chord theorem gives us a way to take the square root of a rectangle's area.

Perpendicular chords

Now consider the special case where the two chords are perpendicular, still with one split into parts p, q by the other, which this first splits into parts r, s, with p.q = r.s. If we half-turn the p+q chord about the centre of the circle, it now splits the r+s chord as s+r and the distance between the original and rotated copy is r −s or s −r, whichever is positive. The diameter of the circle then has square (p+q).(p+q) +(r−s).(r−s) = p.p +q.q +r.r +s.s since 2.p.q −2.r.s = 0; so the areas of the squares on the four parts of our two chords add up to the same area as that of the square that bounds the circle; multiply this area by π/4 to get the area of the circle itself. For an illustration, see Catriona Shearer's puzzle that brought this to my attention.

Ptolemy's Theorem

In the diagram above, the ends of the intersecting chords are connected to form a quadrilateral, whose diagonals are the intersecting chords. We've seen how the parts ot the diagonals relate to one another, so now let's look at how the sides and the diagonals relate to one another.

P Q R S p r q s r q s p In the illustration, PS subtends angle q at both R and Q; it thus subtends angle 2.q at the centre of the circle. Likewise: QS subtends r at P and R so 2.r at the centre, QR subtends p at P and S so 2.p at the centre and PR subtends s at S and Q, hence 2.s at the centre. Since these four edges encircle the centre, we find p+q+r+s equal to a half turn. For the diagonals, PQ subtends q+r on one sie (at R) and p+s on the other (at S), while RS subtends s+q on one side (at Q) and p+r on the other (at P); in each case, of course, the angles on the two sides add up to p+q+r+s = turn/2 again. If the circle's radius is h, we can use the sine result above to obtain the lengths of the sides and diagonals as:

If we multiply opposite sides of our quadrilateral, giving a product for each pair, and add the results we get PS.QR +PR.QS = 4.h.h.(Sin(q).Sin(p) +Sin(s).Sin(r)) in which we can substitute Sin(s) = Sin(p+q+r), as p+q+r+s = turn/2. When we expand out the sines and cosines of sums of angles in terms of the sines and cosines of the angles added, we obtain

(PS.QR +PR.QS)/h/h/4
= Sin(q).Sin(p) +Sin(p+q+r).Sin(r)
= Sin(p).Sin(q) +(Sin(p+q).Cos(r) +Cos(p+q).Sin(r)).Sin(r)
= Sin(p).Sin(q) +(Sin(p).Cos(q) +Cos(p).Sin(q)).Cos(r).Sin(r) +(Cos(p).Cos(q) −Sin(p).Sin(q)).Sin(r).Sin(r)
= Sin(p).Sin(q).Cos(r).Cos(r) +Sin(p).Cos(q).Cos(r).Sin(r) +Cos(p).Sin(q).Cos(r).Sin(r) +Cos(p).Cos(q).Sin(r).Sin(r)
= (Sin(p).Cos(r) +Cos(p).Sin(r)).(Sin(q).Cos(r) +Cos(q).Sin(r))
= Sin(p+r).Sin(q+r)
= RS.PQ/h/h/4

whence PS.QR +PR.QS = RS.PQ.

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