Geometry: # Similar triangles Here [1] we find five key scenarios where similar triangles are found. Top left: the intercept theorem, where two non parallel lines AC and AB are crossed by two parallel lines BC and ED, forming similar triangles ABC and AED. Centre left: a variation of the intercept theorem, where the two parallel lines GI and HF cross the non-parallel two (HI and FG) but at different sides of their intersection point J. Top right: Two non-parallel lines NK and ML meeting at O are crossed by a circle, making similar triangles OMN and OLK. Bottom left: two segments of length 2a and 2b meet at their midpoints, forming two similar triangles with parallel and equal sides c. Bottom right: two segments of length (a+b) cross at their a-b separation, forming two similar triangles, and with the four edges forming a cyclical quadrilateral. [1] https://www.geogebra.org/m/gzbm9twg # Area of triangle = perimeter × radius/2 The area of a triangle is equal to p×r∕2, where p = perimeter and r = radius of its inscribed circle. [Geogebra link:] https://www.geogebra.org/m/bphzn6eh # Visual representation of the six trig functions In this diagram [1] we can see the six trigonometric functions all at once. Firstly, it shows how sine (vertical) and cosine (horizontal) are the sides of the right rectangle with hypotenuse = 1. The tangent is just the tangent segment from our chosen point in the unit circle to the horizontal axis. The cotangent is also tangent, but from the point to the vertical axis. Both are tangent and complementary, and they form a hypotenuse where the sides are the secant (horizontal) and the cosecant (vertical). The horizontal pair, cosine and secant, are reciprocal, and so they are the sine and the cosecant (both vertical). [1] https://www.geogebra.org/m/jxufjfsp # Pick's theorem Area of a simple (non-intersecting) polygon with vertices on nodes of a square grid. First, divide all by the length of the grid cell (L), so all vertices have integer coordinates. The area is then i + b∕2 - 1, where i = interior nodes and b = boundary nodes. Then multiply by L to get the first area. # Angles in a circle In a cercle, let's fix a chord (or an arc). This chord can be seen 1) from the centre, 2) from the circle, 3) from the circumference, and 4) from the exterior of the circle. The angle in each perspective is called 1) central, 2) interior, 3) inscribed, and 4) exterior. The inscribed angle is always the same if we go around the circumference, as long as we stay on the same side of the chord. The central angle is twice the inscribed. For the interior angle, we connect the edges of the chord (A and B) with our interior point P and form two new points (A' from AP and B' from BP). The original chord AB and the new chord A'B' have a central angle each, and we take the mean of the two. This mean is the interior angle. For the exterior angle, we proceed similarly: connect the edges of our chord (A and B) with our exterior point P, to find two new points (A' from AP and B' from BP). The original chord AB and the new chord A'B' have a central angle each, and we take, instead of the mean, the big central angle minus the small one, and divide by two, which gives the exterior angle. # Angle of a chord along a diametre Let α be the central angle to a given chord. The chord and two radii define an isosceles triangle. If we now move our point of view along one of these radii a distance d, so that we shift from the centre a factor f=d∕r, the angle when looking at the same chord is now β. The relation between β and α is tanβ=sinα∕(cosα - f). The proof is elemental if drawing a right triangle with the same angle α. # Morley's trisector theorem Take any triangle and trisect their three inner angles. The intersections of adjacent trisectors will form an equilateral angle, called Morley triangle. https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem#/media/File:Morley_triangle.svg # Medial triangle Consider a triangle ABC with sides a,b,c. Take their midpoints Ma,Mb,Mc and join them. They will form the medial triangle, which is similar to triangle ABC. The centroids of both triangles are the same point. # Angle bisector theorem Consider a triangle ABC, and we locate ourselves in the perspective of A, seeing B to our left and C to our right. Bisect angle A, so that side a (in front of us), is divided into a_c (left) and a_b (right). The theorem states that a_c/a_b = c/b. Intuitively, front_left/front_right = left/right. Remarkably, if proving this theorem in the complex plane, the imaginary part brings this theorem, but the real part brings another, giving the length of the angle bisector, b_a, where b_a² = b·c - a_b·a_c. # A useful count: 2, π, and 4 Consider a circle C, its (bigger, outer) circumscribed square CS and its (smaller, inner) inscribed square IS. We can associate IS=2, C=π and CS=4. Then, we get the area ratios C/CS = π/4, C/IS = π/2 and of course IS/CS = 1/2. # Adjacent and vertical angles Let two lines cross at a point, so that four angles can be considered. We call adjacent angles to those who are neighbours. We call vertical angles to those who aren't neighbours. Then, vertical angles are equal, whereas adjacent angles are supplementary, i.e. their sum is 180° (complementary angles sum 90°). # The tangent formula Consider a circle with centre C and an outer point P. Draw the two tangents to the circle through P, touching the circle at A and B. The radii CA and CB divide the circle in two unequal sectors, one with angle β and another with angle γ=360-β, with β>γ. Then, the angle APB = (β-γ)/2. # Intersecting secants theorem Two lines cross at O. Then, we draw any circle with centre C that crosses them in four points A,B,A',B'. The angle AOB=A'OB' is the arithmetic mean of the angles ACB and A'CB'. # Beyond the intercept theorem: cutting two crossing lines with a circle The intercept theorem (known as Thales' theorem in Spain) is only half of the fun. Let's see why. Take two lines crossing at O. Now, cross these two lines with two parallel lines, one creating A,B and the other A',B'. Thales tells us that the the triangles OAB and OA'B' are similar. Now, instead of cutting the two initial lines with a pair of parallel lines, cut them with \*any\* circle, getting the 4 points A,B,A',B'. Triangles OAB and OA'B' are also similar. # The Euler line Consider a triangle with sides a,b,c; vertices A,B,C; medians ma,mb,mc; altitudes ha,hb,hc; perpendicular bisectors pa,pb,pc and angle bisectors ba,bb,bc. The medians cross at the centroid G, the altitudes at the orthocentre H, the perpendicular bisectors at the circumcentre O, and the angle bisectors at the incentre I. Remarkably, G,H,O are collinear along what is called the Euler line. Moreover, G is always between H and O, and HG=2GO. In other words, H\--G-O. The nine-point centre also belongs to this line. # The nine-point centre Take any triangle with sides a,b,c; altitudes ha,hb,hc (defining the orthocenter O); vertices A,B,C; midpoints Ma,Mb,Mc; feet Fa,Fb,Fc; and Pa,Pb,Pc being midpoints of OA,OB,OC. Then, these 9 points A,B,C,Ma,Mb,Mc,Fa,Fb,Fc,Pa,Pb,Pc belong to the same circle, called the nine-point centre. https://www.geogebra.org/m/dcpzdk8p # Simplex as the generalisation of the triangle and the tetrahedron A simplex is the generalisation of the triangle to other dimensions. Thus, a (0,1,2,3,4)-dimensional simplex is a (point, line segment, triangle, tetrahedron, etc). I wonder if the app, SimpleX, is based on multi-dimensional triangles. # Viviani's theorem A diamond of geometry. Consider an equilateral triangle ABC and an interior point P. From P, take the shortest paths to each side: a, b and c to sides BC, CA and AB, respectively. Then, a + b + c = h (the triangle's altitude). The converse also holds. An interesting extension for parallelograms: the sum of the four distances from any interior point is constant (converse also true). Returning to the triangle version, ternary graphs are plotted in an equilateral triangle where each side is an axis, and where the sum of the three plotted variables is a constant, like colour (chromaticity diagram, with r+g+b=colour), (diploid) allele frequencies (de Finetti diagram, with aa+Aa+AA=1) or soil texture (salt+silt+clay percentages = 100%). This allows to represent three coordinates in two dimensions. # Menelaus' and Ceva's theorems Take a triangle ABC. Now, consider the lines defined by the three sides AB, BC and CA, and place a point on each line: P, Q and R, respectively. Next, go over the vertices of the triangle making a stopover at each corresponding new point. This means that our whole trip is A, P, B, Q, C, R, A. Using this order, built three consecutive fractions and multiply them, like this: AP/PB · BQ/QC · CR/RA = r. With this in mind: 1) Menelaus' theorem says that if P, Q and R are aligned, r = 1. The converse, which is very useful, says that if r = 1, those three points are aligned. 2) Ceva's theorem says that if AP, BR and CP are cevians (segments connecting a vertex of a triangle with its opposide side) and they meet at a point O, then r = 1. The converse, also very useful, says that if r = 1, the three cevians are concurrent (or exceptionally, parallel). In summary: when having three aligned points crossing a triangle (or its extension), use Menelaus'; when needing to prove alignment, use its converse; if three cevians are concurrent, use Ceva's; and if concurrency (or parallelism) needs to be proven, use its converse. # Ptolemy's theorem and the British flag theorem Two lovely geometry theorems that I was never told about in school or uni. 1) Ptolemy's theorem: take four points A, B, C and D on a circumference (cyclically, i.e. in that order). Then, AB·CD + BC·DA = AC·BD. It is a generalisation of Pythagoras' theorem. 2) The British flag theorem: take a rectangle with cyclical points and take any point P inside or outside of the polygon. Then, AP² + PC² = DP² + PB². Which is another generalisation of Pythagoras' theorem. Not a single teacher took the time to introduce me to these gems. Unforgivable.