Geometry

1Angle sums in triangles and polygons

The interior angles of a triangle sum to 180^\circ, and an exterior angle equals the sum of the two remote interior angles — never the adjacent interior. For an n-sided polygon the interior angles sum to (n-2)\times 180^\circ; each exterior angle of a regular n-gon is \frac{360^\circ}{n}, and the exteriors of any convex polygon sum to 360^\circ whether the polygon is regular or not.

Figure. Right isosceles \triangle ABC with the right angle at B; equal legs are drawn to true length under aspect 2 (\Delta x = \Delta y / 2). Side BC is extended (dashed) past C. The exterior at C is the sum of the remotes at A and B, not the 45^\circ sitting next to it.

How it works

1. Close the triangleThree interiors add to 180^\circ. If two are known, the third is forced.
2. Exterior from remotesAn exterior angle equals the sum of the two interiors that do not touch it — the remote pair.
3. Polygon by trianglesAn n-gon splits into n-2 triangles from one vertex, so interiors sum to (n-2)\times 180^\circ.

2Pythagoras and the triples worth knowing

In a right triangle, \text{hyp}^2 = \text{base}^2 + \text{height}^2. The exam almost never wants you to square and root from scratch: it plants a Pythagorean triple — 3,4,5; 5,12,13; 8,15,17; 7,24,25 — or a multiple of one, and the mark is for recognising it on sight.

Any positive multiple of a triple is still a right triangle: 6,8,10 and 9,12,15 are both 3,4,5 scaled. Spot the common factor first, reduce, then read the hypotenuse off the list.

Figure. A 3-4-5 right triangle with legs to true length under aspect 2: vertical \Delta y = 0.48, horizontal \Delta x = 0.32, so the physical base-to-height ratio is 4:3. The right angle sits at the lower left; the hypotenuse is the slant.

How it works

1. Name the right angleThe side opposite the right angle is the hypotenuse — the only one that can sit alone on the left of Pythagoras.
2. Look for a tripleDivide out a common factor and match against 3-4-5, 5-12-13, 8-15-17, 7-24-25.
3. Scale back upMultiply the looked-up hypotenuse by the factor you divided out.

3Centroid, incentre and circumcentre

Three centres of a triangle earn their own questions. The centroid is where the medians meet; it divides each median in the ratio 2:1, with the longer part toward the vertex — so the vertex-to-centroid stretch is \frac{2}{3} of the median. The incentre is where the angle bisectors meet and is equidistant from the three sides. The circumcentre is where the perpendicular bisectors meet and is equidistant from the three vertices.

Figure. Medians concurrent at the centroid G. On median AM, the brackets mark the 2:1 split — longer toward A, shorter toward M. The drawn ratio is exact: AG:GM = 2:1 in figure coordinates. The incentre and circumcentre are not drawn; each needs a circle the vocabulary cannot make.

How it works

1. Centroid from a medianCut the median 2:1 from the vertex. Vertex-to-G is \frac{2}{3}; G-to-midpoint is \frac{1}{3}.
2. Incentre from the sidesEqual perpendicular distance to every side — that common length is the inradius r.
3. Circumcentre from the verticesEqual distance to every vertex — that common length is the circumradius R.

4Heron's formula, inradius and circumradius

Given three sides a, b, c and no height, the area is Heron's formula: with semi-perimeter s = \frac{a+b+c}{2}, Area = \sqrt{s(s-a)(s-b)(s-c)}. Once the area is known, the two radii drop out of identities — inradius r = \frac{\text{Area}}{s} and circumradius R = \frac{abc}{4\times\text{Area}} — with no further geometry.

Figure. Sides 13, 14, 15 give semi-perimeter s=21. Heron then yields Area =\sqrt{21\cdot 8\cdot 7\cdot 6}=84. From that area, r=\mathrm{Area}/s and R=abc/(4\cdot\mathrm{Area}) drop out — the triangle is to scale; the inscribed and circum circles are not drawn.

How it works

1. Form sAdd the three sides and halve: s = (a+b+c)/2.
2. Heron under the rootMultiply s(s-a)(s-b)(s-c) and take the positive square root.
3. Radii from Arear = \text{Area}/s and R = abc/(4\times\text{Area}).

5Similar triangles and the square on the sides

If two triangles are similar with corresponding-side ratio a:b, their areas are in the ratio a^2:b^2. Corresponding medians, altitudes and perimeters stay in the linear ratio a:b — only area (and every other two-dimensional measure) picks up the square. The trap is reading an area ratio as if it were a side ratio, or taking a square root the wrong way when the question starts from areas.

Figure. Two similar triangles with corresponding bases 3 and 6 — exact ratio 1:2 in figure coordinates (0.24 against 0.48). Heights are 0.28 and 0.56, the same 1:2, so areas scale as 1:4. Linear labels stay with the side ratio; only the area annotations carry the square.

How it works

1. Confirm similarityAAA, proportional sides, or an exam statement — then corresponding sides share one ratio.
2. Linear measuresSides, medians, altitudes, perimeters: all scale as a:b.
3. Area scales as the squareArea ratio = (a/b)^2. From areas back to sides, take the positive square root.

6Central angle, inscribed angle and the semicircle

The angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the remaining circumference. Angles in the same segment — standing on the same arc, on the same side of the chord — are equal. The angle in a semicircle is 90^\circ: an angle standing on a diameter, with its vertex on the circle, is a right angle.

How it works

1. Name the arcCentral and inscribed angles only compare when they stand on the same arc.
2. Apply the factor of twoCentral = 2 \times inscribed. Convert before writing any equation.
3. Semicircle shortcutIf one side of the angle is a diameter and the vertex is on the circle, the angle is 90^\circ.

7Tangent perpendicular to radius, and equal tangents

A tangent to a circle is perpendicular to the radius at the point of contact — so the triangle formed by the centre, the external point and the contact point is right-angled at the contact, and Pythagoras applies. Two tangents drawn from the same external point are equal in length; the line from the centre to that external point bisects the angle between them.

How it works

1. Radius meets tangent at 90°At the point of contact, radius \perp tangent. Mark that right angle before computing lengths.
2. Equal tangents from one pointFrom external point P, tangents touching at A and B satisfy PA = PB.
3. Pythagoras on OPWith radius r and OP = d, each tangent length is \sqrt{d^2 - r^2}.

8Intersecting chords

Two chords AB and CD intersecting inside a circle at P satisfy PA \times PB = PC \times PD. The products of the segments of each chord are equal — not the chords themselves, and not the sums. Write the product identity before substituting numbers; adding the known pieces is the usual wrong move.

How it works

1. Name the four segmentsFrom the intersection, each chord splits into two pieces — four lengths, one unknown.
2. Write the productsPA \times PB = PC \times PD. Same chord's two pieces multiply; the other chord matches.
3. Solve for the unknownOne multiplication and one division. Do not add the segments.