Geometry I · Lines, Angles & Triangles

This is the opening type of the chapter. Do not take it lightly, though, because the same three ideas sit inside every type that follows.

Complementary angles. Two angles whose sum is 90° are complementary, and each is the complement of the other. The complement of x is (90° − x), so the complement of 35° is 55°.

Supplementary angles. Two angles whose sum is 180° are supplementary, and each is the supplement of the other. The supplement of x is (180° − x), so the supplement of 35° is 145°.

Linear pair. When a ray stands on a straight line, it makes two adjacent angles that together fill the straight angle. Their sum is 180°, so a linear pair is always supplementary.

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Vertically opposite angles. Two crossing lines make four angles. The angles that face each other across the crossing point are vertically opposite, and they are always equal. The reason is simple. On one line, x and its neighbour make 180°, so the neighbour is 180° − x. On the other line, the angle opposite x is also 180° minus that same neighbour, which brings us back to x. One crossing gives only two values: x and 180° − x.

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Angle equations. Most questions of this type are one sentence long, such as "an angle is 20° more than twice its complement". Write the angle as x and its complement as (90° − x), then turn the sentence into an equation word by word: x = 2(90° − x) + 20°. Solve it to get the answer.

When Radhika opens a pair of scissors, the two blades cross like two lines. If the blades make 40° on one side, the opposite side is also 40°, and each of the two side gaps is 180° − 40° = 140°.

Counting points, lines and regions. A few questions on lines are pure counting. Draw n straight lines with no two parallel and no three passing through one point.

• Each pair of lines meets once, so the maximum number of intersection points = ⁿC₂ = n(n − 1)2.
• The k-th line is cut by the earlier k − 1 lines into k pieces, and each piece splits one region into two. So the number of regions grows 2, 4, 7, 11, … and n lines divide a plane into at most n(n + 1)2 + 1 regions.

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Now turn it around and start with points. If n points have no three on one line, they give ⁿC₂ straight lines and ⁿC₃ triangles. When m of the points are on one line, those m points give only one line instead of ᵐC₂ lines and no triangle at all. So lines = ⁿC₂ − ᵐC₂ + 1 and triangles = ⁿC₃ − ᵐC₃. Radhika marks 7 points with 4 of them on one line: she gets 21 − 6 + 1 = 16 lines and 35 − 4 = 31 triangles.

Two lines in a plane that never meet are parallel (written l ∥ m). A line that cuts two or more lines is a transversal. When a transversal cuts two parallel lines, it makes eight angles, four at each crossing, and three simple rules connect all of them. Number the angles 1 to 8 as in the figure:

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So one angle at a crossing tells you all eight. Four of them equal x and the other four equal 180° − x. In every parallel-line question, you only decide which of the two values the asked angle takes.

The bent-line (zig-zag) rule. Suppose a point E lies between two parallel lines and is joined to a point on each line. Then the angle at E equals the sum of the two angles that the joining lines make with the parallel lines, on the side facing E. To see why, draw a third line through E parallel to both. It splits the angle at E into two parts, and each part is an alternate angle of one of the given angles.

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When the point lies outside the two lines, the same construction gives a difference instead of a sum. The angle at the outside point = (bigger angle) − (smaller angle), both measured on the same side.

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Inside a triangle. If DE ∥ BC in △ABC, then ∠ADE = ∠ABC and ∠AED = ∠ACB, because they are corresponding angles. This small fact is where similarity begins, and Types 7 and 8 depend on it.

Radhika's notebook has parallel ruled lines. A slanting pencil line makes 70° with one ruled line, so it makes 70° with every other ruled line too (corresponding angles), and the angle on the other side of the pencil line is 110°.

No other fact in this chapter is used as often as this one.

Why is it true? Take △ABC and draw a line through A parallel to BC. Three angles now sit on this straight line at A. The left one equals ∠B and the right one equals ∠C, both as alternate angles, and the middle one is ∠A itself. Angles on a straight line make 180°, so ∠A + ∠B + ∠C = 180°. So Type 1 and Type 2 have just worked together.

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A few results follow at once:

• A triangle can have at most one right angle or obtuse angle. Two right angles would already use up the whole 180°.
• In a right-angled triangle the other two angles are complementary: ∠B + ∠C = 90°.
• In an equilateral triangle each angle is 180°3 = 60°.
• If one angle is greater than the sum of the other two, it is more than 90° and the triangle is obtuse. If it equals the sum of the other two, it is exactly 90°.

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How the exam writes it

• Ratio: angles in the ratio a : b : c ⇒ one part = 180°a + b + c.
• Expressions: angles such as (4x − 20)°, (3x + 10)° and (2x + 10)°. Add them, set the sum equal to 180° and find x.
• Differences: "P − Q = 15°, Q − R = 30°". Write all three angles in terms of one of them.
• Equal multiples: "2∠A = 3∠B = 6∠C". Put 2A = 3B = 6C = k. Then A = k2, B = k3, C = k6, so the angles are in the ratio 12 : 13 : 16 = 3 : 2 : 1.
• Angles in AP: the middle angle is always 60°, because (a − d) + a + (a + d) = 3a = 180°.

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Radhika's triangle has its angles in the ratio 1 : 2 : 3. Six parts make 180°, so one part is 30° and the angles are 30°, 60° and 90°. A small ratio, and it has given her a right-angled triangle.

Extend one side of a triangle, say BC beyond C to a point D. The angle ∠ACD formed outside the triangle is an exterior angle. The two angles of the triangle that are not next to it, ∠A and ∠B, are its interior opposite angles.

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The proof takes two lines. ∠ACD + ∠ACB = 180° (linear pair) and ∠A + ∠B + ∠ACB = 180° (angle sum). Both are 180° minus the same ∠ACB, so ∠ACD = ∠A + ∠B. In many questions this saves a whole step, because you never need the third angle.

Two more results:

• The three exterior angles of a triangle, one at each vertex, add up to 360°.
• An exterior angle is always bigger than each of its interior opposite angles.

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Isosceles triangles. If AB = AC, the angles opposite these sides are equal: ∠B = ∠C. The converse is also true: if two angles are equal, the sides opposite them are equal. The angle between the equal sides (∠A) is the vertical angle, and ∠B = ∠C = 180° − A2 are the base angles.

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Angle chasing. Many questions mark equal lengths on a figure, such as XA = YA, CD = AC or BQ = PQ. Each pair of equal sides gives you a pair of equal angles. Then apply the angle sum and the exterior angle rule, one after the other, until you reach the asked angle. Go slowly, and write every angle on the figure the moment you find it.

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The five-pointed star. Draw a star by extending the sides of a pentagon, or by joining every second corner of five points. The five tip angles of the star add up to 180°. Apply the exterior angle rule twice: it moves two of the tip angles next to a third one, so the five tips together form the three angles of a single triangle. This holds for any star of this shape, not only a regular one. If four tips are 30°, 40°, 35° and 50°, the fifth is 180° − 155° = 25°.

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In Radhika's isosceles triangle the vertical angle is 40°. Each base angle is 180° − 40°2 = 70°, and the exterior angle at a base vertex is 180° − 70° = 110°, which is 40° + 70°, just as the rule says.

The arrowhead (dart) rule. Take a quadrilateral ABCD whose interior angle at C is reflex, so the shape points inward at C. Then the ordinary angle ∠BCD, measured on the outside, equals ∠A + ∠B + ∠D. The four interior angles add up to 360° and the interior angle at C is 360° − ∠BCD, so ∠A + ∠B + ∠D + 360° − ∠BCD = 360°. If ∠A = 40°, ∠B = 25° and ∠D = 35°, then ∠BCD = 100°.

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Can any three lengths make a triangle? No. Try sticks of 2 cm, 4 cm and 7 cm. Lay the 7 cm stick down, and the other two together reach only 6 cm, so their ends can never meet. This is the triangle inequality:

It follows that the difference of any two sides is less than the third side.

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You do not have to test all three sums. One test is enough: (sum of the two smaller sides) > (largest side). Since 3 + 4 > 6, the lengths 3, 4, 6 make a triangle. If the sum is exactly equal, as in 3 + 5 = 8, the three points lie on one straight line and there is no triangle.

Range of the third side. If two sides are a and b, the third side x must satisfy

With sides 5 and 9 this gives 4 < x < 14. Both ends are strict: the sign is "<", not "≤".

How many integer values? Count the integers strictly between |a − b| and a + b. When a and b are whole numbers, the count is always 2 × (smaller side) − 1. For 5 and 9 it is 2 × 5 − 1 = 9, the values 5 to 13.

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Side–angle relation. In a triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. If ∠A : ∠B : ∠C = 5 : 3 : 4, the shortest side is opposite B (that is CA) and the longest is opposite A (that is BC).

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Perimeter bound. Each side is less than the sum of the other two, so every side is less than half the perimeter. In a triangle of perimeter 52, every side is less than 26.

Counting triangles. Some questions ask how many triangles have integer sides and a given perimeter, or fix the product of two sides. List the cases in order, smallest side first, and test each one with the triangle inequality. Take care not to count the same triangle twice in a different order.

Radhika has two sticks of 4 cm and 9 cm. The third stick must be longer than 5 cm and shorter than 13 cm, so in whole centimetres she can use 6, 7, …, 12. That is 7 choices, and 2 × 4 − 1 = 7.

Two triangles are congruent (≅) when one can be placed exactly on the other: the same shape and the same size. Then all six parts match, the three sides and the three angles. Their perimeters, areas, medians, altitudes and inradii are all equal as well.

To prove congruence, though, we do not need all six parts. Any one of these five sets is enough:

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Two sets that do not work:

• AAA: three equal angles give the same shape but not the same size. Think of a small and a big equilateral triangle. That is similarity, not congruence.
• SSA: two sides and an angle that is not between them can give two different triangles. RHS is the one special case of this that works, because there the angle is 90°.

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The order of the letters matters. △ABC ≅ △PQR means A ↔ P, B ↔ Q, C ↔ R, so AB = PQ, BC = QR, CA = RP, ∠A = ∠P and so on. If the question writes △ABC ≅ △RPQ, then A ↔ R, B ↔ P, C ↔ Q, and AB = RP, BC = PQ, CA = QR. Write the two names one below the other and read the matching pairs straight down.

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CPCT stands for Corresponding Parts of Congruent Triangles (they are equal). This is how congruence finds a length or an angle: first prove two triangles congruent, then carry the needed side or angle across.

Where it hides in a figure. "AD = AE and ∠BAD = ∠CAE" gives △ABD ≅ △ACE. A median produced, with two perpendiculars dropped on it, gives two congruent right triangles at the midpoint. A point on an angle bisector with perpendiculars to the arms gives equal perpendiculars. "AB = DB, AC = DC" is a kite, and △ABC ≅ △DBC by SSS.

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Radhika cuts two triangles out of paper, both with sides 5 cm, 6 cm and 7 cm. However she turns or flips one of them, it covers the other exactly, so they are congruent by SSS. Both have the same perimeter of 18 cm and the same area.

Congruent triangles are exact copies. Similar triangles (~) are like a photo and its enlargement: the same shape, but not necessarily the same size. In similar triangles the corresponding angles are equal and the corresponding sides are in one common ratio.

As with congruence, the order of the letters tells you which side matches which.

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How to prove similarity (any one is enough):

• AA (or AAA): two angles equal. The third angle is then equal by itself, and this is the test you will use most.
• SSS: all three sides in the same ratio.
• SAS: two sides in the same ratio and the angle between them equal.

Every length scales by k. If the sides are in the ratio k, then the perimeters, and the corresponding altitudes, medians, angle bisectors, inradii and circumradii, are all in the same ratio k. So "perimeters 30 and 45" tells you straight away that the sides are in the ratio 30 : 45 = 2 : 3.

Basic Proportionality Theorem (Thales). A line drawn parallel to one side of a triangle divides the other two sides in the same ratio. In △ABC with DE ∥ BC (D on AB, E on AC):

The reason is that DE ∥ BC gives ∠ADE = ∠B and ∠AED = ∠C, so △ADE ~ △ABC by AA. Be careful here: DE is compared with BC through AD : AB (the whole side), never through AD : DB.

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Midpoint theorem. If D and E are the midpoints of AB and AC, then DE ∥ BC and DE = 12 BC. It is BPT with the ratio 1 : 1. Joining all three midpoints gives a small triangle whose perimeter is half that of the big one.

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A hidden similar pair. If a point S on QR makes ∠PSR = ∠QPR, then △PSR ~ △QPR, because ∠R is common to both. From this, PR² = QR × SR. Whenever a question gives one angle equal to another at a shared vertex, look for this pair.

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Radhika's photo is 4 cm × 6 cm, and the enlarged print is 10 cm wide. The scale factor is 104 = 2.5, so the height becomes 6 × 2.5 = 15 cm. Similar triangles work the same way: find k once, then multiply every length by it.

Area is length × length, so when every length is multiplied by k, the area is multiplied by k². The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. For the same reason it is also the square of the ratio of their perimeters, altitudes or medians.

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Going the other way, when the areas are given, take the square root to get the ratio of the sides (or of the perimeters or altitudes). Areas 49 : 81 ⇒ sides 7 : 9.

The DE ∥ BC picture. Here △ADE ~ △ABC, so ar △ADE : ar △ABC = AD² : AB². The trapezium DECB is what remains: ar DECB = ar △ABC − ar △ADE. So if ar △ADE : ar DECB = 9 : 16, then ar △ADE : ar △ABC = 9 : 25, which gives AD : AB = 3 : 5 and AD : DB = 3 : 2.

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Midpoint triangle. Joining the midpoints of the three sides makes four congruent triangles. Each has 14 of the area of the original, because the side ratio is 1 : 2 and so the area ratio is 1 : 4.

Triangles with the same height. This is the second tool of the type, and it needs no similarity at all. Two triangles with the same height have their areas in the ratio of their bases. So:

• A median divides a triangle into two triangles of equal area.
• A cevian AD with BD : DC = m : n divides the area in the ratio m : n.
• For a point E on AD, the triangles ABE and ABD share the vertex B and have their bases on AD, so ar △ABE : ar △ABD = AE : AD. Such questions need the rule twice.

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Two triangles on the same base. When they stand on the same base, their areas are in the ratio of their heights. Use this when perpendiculars are dropped on a common side.

Concurrent cevians and weights (mass points). When cevians AD, BE and CF meet at one point O, give each vertex a weight so that the two weights on a side are in the inverse ratio of the parts of that side. If BD : DC = m : n, put weight n on B and weight m on C; the foot D then carries the sum, B + C. Scale the numbers (use the LCM) so that every vertex has just one weight. Then, with w for weight:

• AO : OD = w(D) : w(A), and likewise BO : OE = w(E) : w(B) and CO : OF = w(F) : w(C)
• AF : FB = w(B) : w(A), AE : EC = w(C) : w(A)

Take BD : DC = 2 : 3 and AE : EC = 3 : 4. The first gives w(B) : w(C) = 3 : 2, the second gives w(A) : w(C) = 4 : 3. Make w(C) = 6 in both, so w(A) = 8, w(B) = 9, w(C) = 6. Then w(D) = 15 and w(E) = 14, so AO : OD = 15 : 8 and BO : OE = 14 : 9.

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A quick check without weights, for concurrent cevians: AOOD = AFFB + AEEC. Here 98 + 34 = 158, the same answer. The method needs the three lines to meet at one point, so use it only when the question says so (two cevians crossing at O, with the third drawn through O, also qualify).

Radhika has a triangular sheet of area 48 cm². She folds it along the line joining the midpoints of two sides. The folded flap is similar to the sheet with half the sides, so its area is 14 × 48 = 12 cm².

In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: c² = a² + b². The hypotenuse is the side opposite the 90° angle, and it is always the longest side.

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Pythagorean triplets. These are sets of whole numbers with a² + b² = c². Learn them by heart, because then you recognise the answer instead of calculating it:

Decimal sides are often a triplet in disguise. With 0.9 and 4, multiply by 10 to get 9 and 40; the triplet is 9-40-41, so the hypotenuse is 4.1. Every multiple of a triplet is again a triplet.

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A triplet from an odd number n: n, n² − 12, n² + 12. For n = 9 this gives 9, 40, 41, and for n = 11 it gives 11, 60, 61.

Sum and difference of the legs. Since (a + b)² + (a − b)² = 2(a² + b²) = 2c², knowing the hypotenuse and the sum of the legs gives the difference at once, and the other way round.

Acute or obtuse? Let c be the longest side.

When the angle is 60° or 120°. For any angle C between sides a and b, the cosine rule gives c² = a² + b² − 2ab cos C. Two special cases are enough for now:

The cosine rule itself is covered properly in Trigonometry.

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Radhika walks 8 m east and then 6 m north. Her distance from the starting point is √8² + 6² = √100 = 10 m. That is the 6-8-10 triplet, so there was nothing to calculate.

Take △ABC, right-angled at A, and drop the altitude AD on the hypotenuse BC. This figure keeps coming back, because it splits the triangle into two triangles, and both are similar to the original: △DBA ~ △ABC ~ △DAC. Each of them has a right angle and shares one acute angle with the big triangle.

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Let BD = p, DC = q and AD = h, as in the figure. The similar triangles give four results:

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You will use the area formula most often. With legs 12 and 16, the hypotenuse is 20 and the altitude is 12 × 1620 = 9.6. With legs 7 and 24, the hypotenuse is 25 and the altitude is 16825 = 6.72.

The areas of these triangles compare in the same way. A small triangle and the big one are similar, so their areas are in the ratio of the squares of their hypotenuses. For △ACD and △BCA the hypotenuses are AC and BC, so ar △ACD : ar △BCA = AC² : BC².

Take Radhika's triangle with legs 9 and 12. The hypotenuse is 15 (3-4-5 times 3). The altitude is 9 × 1215 = 7.2, and the hypotenuse is cut into 9²15 = 5.4 and 12²15 = 9.6. As a check, 5.4 × 9.6 = 51.84 = 7.2².

Some triangles come up so often that their side ratios should be at your fingertips. Then you never need Pythagoras for them.

45°-45°-90° (isosceles right triangle). The two legs are equal, and the sides are in the ratio 1 : 1 : √2. Hypotenuse = leg × √2, and leg = hypotenuse√2. A hypotenuse of 14 gives legs of 7√2 each. The altitude on the hypotenuse is half the hypotenuse.

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30°-60°-90°. The sides opposite 30°, 60° and 90° are in the ratio 1 : √3 : 2. So the side opposite 30° is half the hypotenuse. Read it the other way too: if the hypotenuse is double one side, the angle opposite that side is 30°.

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Equilateral triangle (side a). Its altitude cuts it into two 30-60-90 triangles, so:

• the altitude is also the median and the angle bisector
• from any point inside, the perpendiculars to the three sides add up to h. (Join the point to the three vertices; the three small triangles together make the whole area.)

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Isosceles triangle (equal sides s, base b). The altitude from the top vertex falls on the midpoint of the base, so

Very often the equal side and half the base form a 5-12-13 or 8-15-17 triplet.

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A point on a side of an equilateral triangle. If D is on BC with BD = x, the 60° angle at B gives AD² = a² + x² − ax, the 60° rule from Type 9.

Radhika's set-square has angles 30°, 60° and 90°, and its longest edge is 12 cm. The shortest edge is 6 cm (half of 12) and the middle one is 6√3 cm.

In △ABC, let AD bisect ∠A and meet BC at D. Where exactly does D fall? Not at the midpoint; that would be the median. The bisector divides the opposite side in the ratio of the two sides that make the angle:

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Here is why. △ABD and △ACD have the same height from A, so their areas are in the ratio BD : DC. Also, D lies on the bisector, so it is at the same distance from AB and from AC. Taking AB and AC as bases, the same two areas are in the ratio AB : AC. Both ratios describe the same pair of areas, so they must be equal.

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So, with BC = a:

Each part of BC goes with the side next to it: BD with AB, and DC with AC.

External bisector theorem. The bisector of the exterior angle at A meets BC produced at E. Then

Here E lies outside the segment BC, beyond the end of the shorter of AB and AC, and BE − CE = BC (when AB > AC). With AB : AC = 7 : 4 and BC = 12, we get BE : CE = 7 : 4. The difference of 3 parts is 12 cm, so one part is 4 cm and CE = 16 cm.

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A few facts that come with bisectors

• The internal and external bisectors at the same vertex are perpendicular to each other, since half of 180° is 90°.
• If the bisector of ∠A is also perpendicular to BC, the triangle is isosceles with AB = AC, and D is the midpoint of BC.
• Angle between the bisector and the altitude from the same vertex A: ∠DAE = |∠B − ∠C|2, where AE ⊥ BC.
• Length of the bisector, when it is needed: AD² = AB × AC − BD × DC.

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In Radhika's triangle AB = 6, AC = 9 and BC = 10. The bisector of ∠A cuts BC in the ratio 6 : 9 = 2 : 3, so BD = 4 and DC = 6.

The three internal angle bisectors of a triangle meet at one point, the incentre (I). Every point on an angle bisector is equally far from the two arms of the angle, so I is equally far from all three sides. That common distance is the inradius r, and the circle with centre I and radius r touches all three sides; it is called the incircle. The incentre always lies inside the triangle, whatever kind of triangle it is.

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The angle at the incentre. In △BIC, ∠IBC = B2 and ∠ICB = C2. So
∠BIC = 180° − B + C2 = 180° − 180° − A2 = 90° + A2
In the same way ∠AIC = 90° + B2 and ∠AIB = 90° + C2. In words: the angle at I that faces a side is 90° plus half the angle at the opposite vertex.

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Inradius. Join I to A, B and C. The triangle splits into three triangles, each with height r and with bases a, b and c, so
Area Δ = 12r(a + b + c) = r × s ⇒ r = Δs (s = semi-perimeter)
For a right triangle with legs a, b and hypotenuse c, this becomes r = a + b − c2. For 3-4-5, r = 1; for 5-12-13, r = 2; for 8-15-17, r = 3.

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Excentre: the bisectors of the exterior angles. Extend AB and AC. The bisectors of the two exterior angles at B and C meet at an excentre, which lies outside the triangle. The angle there is
∠BPC = 90° − A2

One internal and one external bisector. The bisector of the internal ∠B and the bisector of the exterior angle at C meet at a point S with
∠BSC = A2
At C the two bisectors are perpendicular to each other, and the exterior angle rule gives the rest.

[[fig:3]]

Learn the three cases together with A = 60°: internal–internal 120°, external–external 60°, internal–external 30°.

Take Radhika's triangle with ∠A = 80°. The internal bisectors from B and C meet at 90° + 40° = 130°, the external bisectors at 90° − 40° = 50°, and an internal with an external at 40°.

A median joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid G, the balancing point of the triangle: a triangle cut from cardboard balances on a pin placed at G. Like the incentre, the centroid always lies inside the triangle.

The 2 : 1 rule. G divides every median in the ratio 2 : 1, counting from the vertex. So if AD is a median, AG = 23AD and GD = 13AD. A median of 15 cm gives AG = 10 cm and GD = 5 cm.

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Areas around the centroid. A median halves the area (Type 8). The three medians together cut the triangle into six small triangles of equal area, each 16 of △ABC. So:

• If L and M are the midpoints of two sides, ar △GLM = 112 ar △ABC. Here LM is half the third side, and △GLM is a quarter of the matching 13-triangle.

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Apollonius theorem. For the median AD on side BC:

It gives the length of a median from the three sides: AD² = 2b² + 2c² − a²4. In a right triangle, the median to the hypotenuse is just half the hypotenuse (Type 15).

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When two medians are perpendicular. If medians AD and BE cross at G at 90°, then △AGB is right-angled at G, with AG = 23AD and BG = 23BE. So AB² = AG² + BG². Many questions are nothing more than this one right triangle. Two more results follow from it:

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Two more facts. The triangle whose sides equal the three medians has 34 of the area of the original. If all three medians are equal, the triangle is equilateral.

Radhika balances a triangular plate on her finger. The balancing point lies 23 of the way down each median from the vertex. On a median of 7.5 cm, that is 5 cm from the vertex and 2.5 cm from the side.

Circumcentre (O). The perpendicular bisectors of the three sides meet at the circumcentre. Every point on the perpendicular bisector of BC is equally far from B and C, so O is equally far from all three vertices: OA = OB = OC = R, the circumradius. The circle through A, B and C, the circumcircle, has its centre at O.

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Where does O lie? That depends on the largest angle:

• acute triangle → inside
• right triangle → at the midpoint of the hypotenuse, so R = hypotenuse2, and the median to the hypotenuse equals R
• obtuse triangle → outside, beyond the longest side

Results to know: ∠BOC = 2∠A (for an acute angle A), and the perpendicular from O to BC bisects ∠BOC, so ∠BOM = ∠A, where M is the midpoint of BC. For any triangle, R = abc4Δ, that is, abc = 4RΔ.

[[fig:1]]

Orthocentre (H). The three altitudes meet at the orthocentre.

• acute triangle → inside
• right triangle → at the vertex of the right angle, because the two legs are themselves altitudes
• obtuse triangle → outside, behind the obtuse angle

The angle at H: ∠BHC = 180° − ∠A. This holds whenever ∠B and ∠C are acute, which covers the usual questions. Here is why. A, the feet of two altitudes and H form a quadrilateral with two angles of 90°, so the other two angles, ∠A and the angle at H, add up to 180°. ∠BHC is vertically opposite that angle at H.

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The feet of the three altitudes form the orthic triangle. Similar triangles give two useful facts here: if BE and AD are altitudes, then △CDE ~ △CAB, so ∠CED = ∠B and ∠CDE = ∠A.

The four centres together.

Euler line. H, G and O always lie on one straight line, and HG : GO = 2 : 1.

[[fig:4]]

Equilateral triangle. All four centres are the same point, so the distance between any two of them is zero. With side a and height h = √32a:
R = 23h = a√3, r = 13h = a2√3, R : r = 2 : 1 and h = 3r. The centre is at a distance r from each side.
The converse also holds: if any two of the four centres coincide, the triangle is equilateral.

Radhika's triangle has sides 6, 8 and 10. It is right-angled, so its circumcentre is the midpoint of the side 10 and R = 5, while its orthocentre is the vertex between the sides 6 and 8.

The pedal triangle. Join the feet D, E, F of the three altitudes and you get the orthic or pedal triangle DEF. Since △CDE ~ △CAB, ∠CDE = ∠A, and in the same way ∠BDF = ∠A. These two angles and ∠EDF together make the straight angle at D, so in an acute triangle the angles of the pedal triangle are 180° − 2A, 180° − 2B and 180° − 2C.

[[fig:5]]

There is one more result hidden here. ∠ADE = 90° − ∠CDE = 90° − A and ∠ADF = 90° − A as well, so the altitude AD bisects ∠EDF. The same is true at E and F. That makes the orthocentre H the incentre of the pedal triangle. Take Radhika's triangle with angles 50°, 60° and 70°: its pedal triangle has angles 80°, 60° and 40°, which again add up to 180°.

Two more results. If AD is the altitude on BC, then Δ = 12 × BC × AD, and R = abc4Δ turns into R = bc2 × AD: the product of the two sides at a vertex, divided by twice the altitude from that vertex. With b = 13, c = 15 and altitude 12 on a = 14, R = 13 × 1524 = 8.125. In an acute triangle with altitudes AD, BE and CF meeting at H, AH × HD = BH × HE = CH × HF. The reason: ∠ADB = ∠AEB = 90°, so A, B, D, E lie on a circle with diameter AB, and the chords AD and BE meet at H. If AH = 6, HD = 4 and BH = 8, then HE = 6 × 48 = 3.