Geometry of Triangles and Circles

Pythagoras theorem and triples

The Pythagoras theorem states that in a right-angled triangle the square of the hypotenuse, the side opposite the right angle, equals the sum of the squares of the other two sides, so c^2 = a^2 + b^2. The hypotenuse is always the longest side. When the sides are given as algebraic expressions or as a ratio, the largest expression must be taken as the hypotenuse, and every side must be positive, so roots that make a side non-positive are rejected after solving the resulting quadratic. When only the perimeter and the hypotenuse are known, the legs can be found without solving for each one separately. Write a + b = P - c from the perimeter and use (a+b)^2 = a^2 + b^2 + 2ab = c^2 + 2ab to obtain the product ab, which gives the area as \frac{1}{2}ab. The same idea works in reverse. If the perimeter and the area are known, then a + b + c = P and ab = 2A, and substituting c = P - (a+b) into c^2 = a^2 + b^2 yields an equation in a + b alone, from which the hypotenuse follows. In a right triangle the two acute angles are complementary, so \tan A \cdot \tan C = 1 and \tan C = \cot A. Therefore \tan A + \tan C = \tan A + \cot A = \frac{1}{\sin A \cos A}, and if \sin A + \cos A is known, squaring it gives 2\sin A \cos A. A right triangle with a leg that is a multiple of an expression can be solved by writing the area as \frac{1}{2}\times(leg)\times(leg). Setting this equal to the given area yields a polynomial equation, which often factorises, and only the positive root that makes every side positive is admissible. Right-angled triangles with integer sides form Pythagorean triples. The common primitive triples are (3,4,5), (5,12,13), (7,24,25), (8,15,17) and (9,40,41), and every multiple of a triple is also a triple. Recognising a triple from the given data often shortens the work, since the hypotenuse and legs can be read off directly.

An isosceles right triangle has legs a and a and hypotenuse a\sqrt{2}, so its perimeter is a(2+\sqrt{2}) and its area is \frac{1}{2}a^2. Once the perimeter is matched to this form, the leg is found by comparing coefficients. The midpoint of the hypotenuse of a right triangle is the circumcentre, so it is equidistant from all three vertices. The distance from this midpoint to the right-angle vertex is therefore half the hypotenuse, and the hypotenuse is twice that distance. In a triangle whose base is divided by the foot of a perpendicular into segments in a fixed ratio, the difference of the squares of the two other sides is proportional to the difference of the squares of the segments, so AB^2 - AC^2 = (AB+AC)(AB-AC) links the ratio of the segments to the base. When one side is changed by a factor and the hypotenuse becomes known, the two unknown legs can be found from two equations, AB^2 + BC^2 = AC^2 and the new relation with the scaled legs, by eliminating one squared leg. For a right triangle with hypotenuse c and legs a > b, the expression (a+b+c)(a+b-c)(a-b+c)(a-b-c) simplifies by grouping into [(a+b)^2 - c^2][(a-b)^2 - c^2], and using a^2 - c^2 = -b^2 reduces it to -4b^2c^2. Recognising such grouping avoids long expansion. When squares are constructed externally on the three sides of a triangle, the total area is the sum of the triangle area and the three squares. Each square's area is the square of its side, so a lateral side of length \sqrt{a^2+h^2} contributes a^2+h^2. For a point D on BC with AD perpendicular to BC, the two right triangles ABD and ACD give BD^2 - CD^2 = AB^2 - AC^2 = (AB-AC)(AB+AC). Since BD^2 - CD^2 = (BD-CD)(BD+CD), a known difference of segments yields their sum and hence the base. The length of a median to the hypotenuse of a right triangle, and of the median from the acute vertex, can be found from AQ^2 = AB^2 + (BC/2)^2 when the right angle is at B. Two such median relations give two equations in two unknown legs, so both pieces of information are needed to fix the area. In an isosceles triangle with AB = AC, a given base and height determine the sides, but the base may also be found from the area by writing BC = 2x and h = \sqrt{AB^2 - x^2}. Solving the area equation produces two possible bases, and a condition such as the height being greater than half the base decides which one is valid. For the isosceles case with a cevian AD to the base, AB^2 = AD^2 + BD \cdot DC, so the square of a side equals the square of the cevian plus the product of the base segments. Substituting the given lengths gives the required side directly.

Altitude to hypotenuse in right triangle

When the perpendicular is drawn from the right-angle vertex to the hypotenuse, it divides the hypotenuse into two segments p and q, and the altitude h satisfies h^2 = pq, so the altitude is the geometric mean of the two segments. The same triangle gives a second set of relations. Each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg, so AB^2 = AD \cdot AC and BC^2 = DC \cdot AC. Two different expressions for the area give the altitude directly. Since the area equals \frac{1}{2}\timesleg\timesleg and also \frac{1}{2}\timeshypotenuse\timesaltitude, the altitude is h = \frac{\text{leg} \times \text{leg}}{\text{hypotenuse}}. Using the same area equality with a different base gives another altitude. Taking the hypotenuse as the base, the altitude from the right-angle vertex is h = \frac{ab}{c}, and taking a leg as the base gives the other leg itself, which is why the method works for any vertex. If a leg and the altitude to the hypotenuse are known, the product of the legs equals the product of the hypotenuse and the altitude, so ab = ch. The circumradius of a right triangle is half the hypotenuse, because the hypotenuse is a diameter of the circumcircle. Hence the hypotenuse is twice the circumradius, and combining this with the area formula using the altitude gives the area directly. The product of the two segments of the hypotenuse equals the square of the altitude, so with AD^2 = BD \cdot DC the altitude can be found whenever the segments are known. Given the sides, the hypotenuse and the altitude to it are linked by h = \frac{ab}{c}, and the same altitude gives the foot of the perpendicular, which in turn determines the two segments of the hypotenuse via AB^2 = AD \cdot AC and BC^2 = DC \cdot AC. When the two segments are fixed by a perpendicular, the squares of the legs are AP = \frac{AB^2}{AC}, PC = \frac{BC^2}{AC} and BP = \frac{AB \cdot BC}{AC}. The areas of the two small triangles formed with the foot are then found from these lengths, and their difference is a straightforward subtraction. A circle centred at a vertex with radius equal to the altitude to the hypotenuse meets the opposite leg at a distance from the vertex equal to that altitude, so the remaining part of the leg is found by subtraction. Because AB \cdot BC = AC \cdot p, the altitude from B onto AC can be found from any two sides, even when the third side is an irrational number such as 2\sqrt{11}, which is obtained first from the Pythagoras relation. Finally, when the two legs are \sqrt{3}p and p, the hypotenuse becomes 2p and the altitude is \frac{p \cdot \sqrt{3}p}{2p}, which simplifies to a multiple of \sqrt{3}p, showing that the altitude formula is always taken with the hypotenuse, not the leg, in the denominator.

Chord properties of circles

A chord is a line segment joining two points on a circle, and a few chord properties are used constantly in CDS problems. The perpendicular drawn from the centre to a chord bisects the chord, and conversely the line from the centre to the midpoint of a chord is perpendicular to it. Equal chords are equidistant from the centre, and the perpendicular bisector of any chord passes through the centre. If a chord subtends a central angle \alpha in a circle of radius r, its length is L = 2r\sin(\alpha/2). For a unit circle this becomes 2\sin(\alpha/2), so a chord subtending 2\theta at the centre has length 2\sin\theta. To find the distance d from the centre to a chord of length L, use the right triangle formed by the radius, half the chord and the perpendicular: d = \sqrt{r^2 - (L/2)^2}. For example, a chord of length 24 in a circle of radius 13 lies at distance \sqrt{169-144} = 5 from the centre. If an angle inscribed in a circle is a right angle, the chord opposite it is a diameter, which is the converse of Thales' theorem. So the circumradius of a right triangle is half its hypotenuse. For legs 6 and 8, the hypotenuse is 10 and the circumradius is 5.

• Tangent to inner circle — for concentric circles, a chord of the outer circle that touches the inner circle is bisected at the point of tangency. If the half-chord is h, then R^2 - r^2 = h^2, so the area between the circles is \pi h^2.
• Sagitta of an arc — if a circular arc has span c and height (sagitta) h at its middle, the radius is R = \dfrac{(c/2)^2 + h^2}{2h}. For span 12 and height 2, R = \dfrac{36+4}{4} = 10.
• Midpoint of a minor arc — if P is the midpoint of the minor arc AB, then OP is perpendicular to AB and passes through the midpoint of AB, so the centre, the chord midpoint and P are collinear. A chord equal to the radius subtends 60^\circ at the centre.
• Intersecting chords theorem — if two chords AB and CD meet at P, then AP \times PB = CP \times PD. The product of the segments of one chord therefore equals the product for the other, so no extra data is needed when one product is known.

Similar triangles: perimeter and area ratios

Similar triangles have equal corresponding angles and proportional corresponding sides with a constant scale factor k. Every linear length, including perimeter, altitude, median and circumradius, is multiplied by k, while every area is multiplied by k^2. So if the perimeters are in ratio p:q, then the sides, and sums of corresponding sides, are also in ratio p:q.

• Perimeter ratio — equal to the ratio of corresponding sides, so it is linear, not squared.
• Area ratio — equal to the square of the ratio of corresponding sides or heights. For example, areas 9:16 give heights 3:4.
• Equal vertical angles in isosceles triangles — two isosceles triangles with the same apex angle are similar, so their heights are in the ratio of their sides.
• Parallel sides in a trapezium — if AB \parallel DC, the triangles formed by the diagonals at their intersection are similar by the AA criterion, since alternate angles are equal. Their area ratio is the square of the ratio AB : DC.
• Tiling with equal triangles — if n^2 equal equilateral triangles form a larger equilateral triangle, the side of the larger one is n times the side of the small one, and its area is n^2 times. The height of an equilateral triangle of side s is \dfrac{\sqrt{3}}{2}s.
• Line parallel to base — if DE \parallel BC in triangle ABC, then triangle ADE is similar to ABC. If ADE is half the area of ABC, then AD/AB = 1/\sqrt{2}, so BD/AB = 1 - 1/\sqrt{2}. A part of a segment must lie between 0 and 1, which is a useful check.
• Areas of equilateral triangles — the area of an equilateral triangle of side a is \dfrac{\sqrt{3}}{4}a^2, so adding areas corresponds to adding squares of sides: a_Z^2 = a_X^2 + a_Y^2.
• Triangle to regular hexagon — cutting equal corners of side a/3 from an equilateral triangle of side a leaves a regular hexagon of side a/3. Its area is \dfrac{3\sqrt{3}}{2}\left(\dfrac{a}{3}\right)^2 = \dfrac{\sqrt{3}}{6}a^2, so the triangle to hexagon area ratio is 3:2.
• Trapezium and common height — the area of a trapezium is \frac{1}{2}(\text{sum of parallel sides}) \times \text{height}. When two figures share the same height, their area ratio equals the ratio of their sums of parallel sides, so compare figures by splitting them into trapezia or triangles with a common height.

Angles in a circle (inscribed angle)

The inscribed angle theorem states that an angle at a point on the circumference equals half the central angle subtending the same arc. Consequently, angles subtended by the same chord at points on the same arc are equal, and an angle in a semicircle is a right angle. When the vertex lies on the arc opposite to the arc being subtended, use the arc that does not contain the vertex. For instance, if the central angle is 90^\circ, the inscribed angle on that arc is 45^\circ.

• Same arc, same angle — in an equilateral triangle ABC inscribed in a circle, \angle BAC = 60^\circ. Any other point D on the same side of BC gives \angle BDC = 60^\circ, whether or not DBC is scalene.
• Diameters and the centre — if two diameters intersect, their intersection is the centre. Triangles formed with two radii and an angle of 60^\circ are equilateral. Angles at the centre along a straight line sum to 180^\circ, so \angle BPD = 180^\circ - \angle APD.
• Reflection and angles of elevation — when an object is viewed reflected in a still water surface, the image is as far below the surface as the object is above it. Use the image point to form the right triangle, so that \tan\theta equals the vertical distance to the image divided by the horizontal distance.
• Sectors and cyclic quadrilaterals — an angle of a sector larger than a semicircle is a reflex angle, which exceeds 180^\circ, so it cannot be less than a right angle. In a cyclic quadrilateral, equal opposite sides subtend equal arcs, and equal arcs give equal diagonals.

Externally touching circles

Two circles touch externally when the distance between their centres equals the sum of the radii, d = R + r. They touch internally when d = R - r. Given the sum of the radii and the sum of their squares, the difference follows from (R-r)^2 = 2(R^2 + r^2) - (R+r)^2. For example, if R + r = 10 and R^2 + r^2 = 58, then (R-r)^2 = 116 - 100 = 16, so R - r = 4, with radii 7 and 3. The area of a circle is \pi r^2, so sums of areas give sums of squares of radii after dividing by \pi.

• Three mutually touching circles — the centres of three pairwise externally touching circles form a triangle whose sides are the sums of the radii in pairs. Adding the three distances gives twice the sum of the radii, so the sum of the radii is half the perimeter of the triangle of centres.
• Identical touching circles — three identical circles of radius r touching each other have centres forming an equilateral triangle of side 2r. The enclosed region is the triangle minus the three 60^\circ sectors, which together form a semicircle. The area is \dfrac{\sqrt{3}}{4}(2r)^2 - \dfrac{\pi r^2}{2}.
• Tangency conditions — for a small circle touching one circle internally and another externally, write the distance from its centre to each centre. Then eliminate the coordinates of the centre to obtain a linear relation between the radii.

Interior and exterior angles of regular polygon

For a regular polygon with n sides, each exterior angle is e = \dfrac{360^\circ}{n} and each interior angle is 180^\circ - e = \dfrac{(n-2)180^\circ}{n}, so the two always sum to 180^\circ. The ratio of the interior angle to the exterior angle is \dfrac{n-2}{2}.

• Difference of the two angles — the difference is (180^\circ - e) - e = 180^\circ - 2e. Setting this equal to a given value gives e, and then n = 360^\circ/e. For a difference of 144^\circ, e = 18^\circ and n = 20. For a difference of 120^\circ, e = 30^\circ and n = 12.
• Possible exterior angles — an exterior angle is valid only if 360^\circ/e is a whole number n \ge 3, so e must divide 360^\circ exactly.
• Inradius of a regular polygon — for side a, the inradius is r = \dfrac{a}{2}\cot\dfrac{\pi}{n}, and the diameter of the inscribed circle is a\cot\dfrac{\pi}{n}. For a unit side and n = 15, \dfrac{\pi}{15} = 12^\circ, so the diameter is \cot 12^\circ. For a square of side 2, \cot 45^\circ = 1 gives diameter 2.

Cyclic quadrilateral opposite angles

A quadrilateral is cyclic when all four of its vertices lie on one circle. Its key property is that each pair of opposite angles is supplementary, so A + C = 180^\circ and B + D = 180^\circ. Therefore each half-sum in a trigonometric expression such as \sin\left(\frac{A+C}{2}\right) equals 90^\circ, and its sine is 1. A cyclic parallelogram must have equal and supplementary opposite angles, so every angle is 90^\circ and the figure is a rectangle. If a cyclic quadrilateral has a pair of parallel sides, it is an isosceles trapezium, so its non-parallel sides are equal and the angles on each leg are supplementary: \angle A + \angle D = 180^\circ as co-interior angles, and \angle A + \angle C = 180^\circ as opposite angles. When a parallelogram has a circle through three of its vertices that meets a produced side at a fourth point, the four points form an isosceles trapezium, so the legs of that trapezium are equal and the segment joining the fourth point to a vertex equals a side of the parallelogram. The internal angle bisectors of a cyclic quadrilateral meet to form another cyclic quadrilateral, because the opposite angles of the inner figure add to 180^\circ. When two opposite sides of a cyclic quadrilateral are produced to meet at a point E outside the circle, the power of a point gives EB \cdot EA = EC \cdot ED. This makes \triangle EBC \sim \triangle EDA with B matching D and C matching A, not \triangle EAD. The sum of the angles \angle CBE and \angle DAE equals \angle A + \angle D of the quadrilateral, which is not necessarily 180^\circ.