Shortcuts & Speed Methods

The Bend Rule for Parallel Lines

Most SSC geometry questions turn on one fact: a bend between parallel lines, a centre angle, a chord distance or a tangent length. If you spot that fact in the first few seconds, the sum rarely needs more than three lines. Draw a rough figure for every question, even when one is printed, and mark everything you are given on it.

When to use it: two parallel lines with a bent path between them (an arrowhead or a zigzag), and the angle at the bend is asked.

Fast way: when the bend points towards the same side as A and C, as in the figure, the angle at the bend is the sum of the two angles the arms make with the parallel lines.

$$\angle BED = \angle ABE + \angle CDE$$

Why it works: draw a line through E parallel to AB (dashed). It cuts angle BED into two parts, and each part is an alternate angle to one of the given angles.

Reverse case: when the bend points the other way (E sits further right than B and D), the three angles add up to $360^\circ$, so $\angle ABE + \angle BED + \angle CDE = 360^\circ$.

More bends: with a zigzag of several bends between the same two parallel lines, the angles that open to the left add up to the angles that open to the right. In the figure, angles ABE and CDE open to the left and angle BED opens to the right.

Question 1

In the figure, $AB \parallel CD$. If $\angle ABE = 42^\circ$ and $\angle CDE = 37^\circ$, find $\angle BED$.

Answer. The bend points towards A and C, so just add: $42^\circ + 37^\circ$ = 79°.

Time check: about 5 seconds.

When it does not work:

• The two lines must be given as parallel. Lines that only look parallel in a figure don't count.
• The given angles must be the ones between each arm and the part of the line on the side the bend points to. If the question gives the angle on the other side at B, take it away from $180^\circ$ first.
• If the bend points away from A and C, use the $360^\circ$ rule. For example, with $\angle ABE = 130^\circ$ and $\angle CDE = 145^\circ$ on that side, $\angle BED = 360^\circ - 275^\circ = 85^\circ$.

[SSC Tip] The bisectors of two co-interior angles (the two angles between the parallel lines on the same side of a transversal) always meet at $90^\circ$, because the halves add up to half of $180^\circ$.

Polygon Sides from One Angle

When to use it: a regular polygon, with one interior or exterior angle given (or their ratio or difference), and the number of sides, diagonals or the angle sum asked.

Fast way: always go through the exterior angle. It is $180^\circ$ minus the interior angle, and the exterior angles of any polygon add up to $360^\circ$. So $n = \frac{360^\circ}{\text{exterior angle}}$.

• Interior angle $I$ given: $n = \frac{360}{180 - I}$.
• Interior : exterior $= p : q$ given: $n = \frac{2(p + q)}{q}$.
• Interior minus exterior $= k$ given: exterior $= \frac{180 - k}{2}$.
• Sum of interior angles $S$ given: $n = \frac{S}{180} + 2$.
• Number of diagonals: $\frac{n(n - 3)}{2}$.

Values worth knowing:

Question 2

The ratio of each interior angle to each exterior angle of a regular polygon is 13 : 2. How many diagonals does it have?

Answer. The two angles add up to $180^\circ$, so the exterior angle is $\frac{2}{15} \times 180^\circ = 24^\circ$ and $n = \frac{360}{24} = 15$. Diagonals $= \frac{15 \times 12}{2}$ = 90.

Time check: about 15 seconds.

When it does not work:

• If $360$ divided by the exterior angle is not a whole number, no regular polygon has that angle. An interior angle of $130^\circ$ gives an exterior angle of $50^\circ$, and $\frac{360}{50} = 7.2$, so there is no such polygon.
• The each-angle formulas need a regular polygon. For any other polygon, only the sums hold.
• Diagonals use $n - 3$; the angle sum uses $n - 2$. Mixing them is the usual slip.

Worth knowing: the only polygon with as many diagonals as sides is the pentagon: $\frac{n(n - 3)}{2} = n$ gives $n = 5$.

Quadrilateral Facts That Save Time

When to use it: a rhombus, parallelogram, rectangle or trapezium with diagonals, sides or angle bisectors in the question.

Fast way:

• Rhombus. The diagonals cut each other in half at right angles, so each side is the hypotenuse of a right triangle with legs $\frac{d_1}{2}$ and $\frac{d_2}{2}$. In one line, $d_1^2 + d_2^2 = 4a^2$. Look for a triplet in the halves.
• Parallelogram. $d_1^2 + d_2^2 = 2(a^2 + b^2)$, where $a$ and $b$ are two adjacent sides. Adjacent angles add up to $180^\circ$, so the bisectors of two adjacent angles meet at $90^\circ$.
• Rectangle and square. The diagonals are equal. A square of side $a$ has diagonal $a\sqrt{2}$.
• Any quadrilateral ABCD. The bisectors of angles A and B meet at an angle of $\frac{\angle C + \angle D}{2}$.
• Trapezium with $AB \parallel CD$. The line joining the midpoints of the slant sides is $\frac{AB + CD}{2}$, and the diagonals cut each other in the ratio $AB : CD$.

In the rhombus PQRS above, the diagonals are 42 and 40, so the halves are 21 and 20, and the side is 29 (the 20, 21, 29 triplet).

Question 3

The sides of a parallelogram are 6 cm and 13 cm, and one diagonal is 11 cm. Find the other diagonal.

Answer. $d_1^2 + d_2^2 = 2(6^2 + 13^2) = 2(36 + 169) = 410$. So $d_2^2 = 410 - 121 = 289$, and the other diagonal is 17 cm.

Time check: about 20 seconds.

When it does not work:

• $d_1^2 + d_2^2 = 4a^2$ is only for a rhombus (a square is a rhombus too). For a parallelogram with unequal sides, use $2(a^2 + b^2)$.
• Use half of each diagonal for the rhombus side. Forgetting to halve gives a side twice as long.
• The $\frac{\angle C + \angle D}{2}$ rule is for the internal bisectors of two angles next to each other, A and B.

Isosceles Chains: Push One Angle Along

When to use it: equal sides are marked (AB = AC, AD = BD, BD = BC and so on), there is an extra point on a side, and an angle is asked.

Fast way: call the first unknown angle $x$. Every pair of equal sides gives two equal angles opposite them, and every exterior angle of a triangle equals the sum of the two interior opposite angles. Push $x$ round the figure until every angle is written in terms of $x$, then use $180^\circ$.

The classic chain is $AB = AC$, with D on AC so that $AD = BD = BC$.

• $AD = BD$, so $\angle ABD = \angle A = x$.
• $\angle BDC$ is an exterior angle of triangle ABD, so it is $2x$.
• $BD = BC$, so $\angle C = \angle BDC = 2x$.
• $AB = AC$, so $\angle ABC = \angle C = 2x$.
• In triangle ABC, $x + 2x + 2x = 180^\circ$, so $x = 36^\circ$.

Question 4

In triangle ABC, $AB = AC$ and $\angle BAC = 112^\circ$. D is a point on BC such that $AD = BD$. Find $\angle DAC$.

Answer. The base angles are $\frac{180^\circ - 112^\circ}{2} = 34^\circ$ each. Since $AD = BD$, $\angle BAD = \angle B = 34^\circ$. So $\angle DAC = 112^\circ - 34^\circ$ = 78°. Check in triangle ADC: $\angle ADC = 34^\circ + 34^\circ = 68^\circ$, and $68^\circ + 78^\circ + 34^\circ = 180^\circ$.

Time check: about 20 seconds.

When it does not work:

• Equal sides give equal angles opposite them. $AD = BD$ makes the angles at A and B of triangle ABD equal, not the angle at D.
• The exterior-angle step needs the extra point to lie on the side itself, not on the side extended. If D is on BC extended, draw it that way and chase again.

A related fact: in an isosceles triangle, the bisector of the angle between the equal sides is also the median and the altitude. The bisector of the exterior angle at that vertex is parallel to the base.

Congruence: Name the Rule in Five Seconds

When to use it: two triangles with some parts equal, and the question asks which rule makes them congruent, which set of conditions is not enough, or which parts must be equal.

Fast way: learn the five rules and the two non-rules.

SSA (the angle is not between the two sides) and AAA (three angles, no side) are not rules. Equal angles only make triangles similar: the same shape, but any size.

Reading the letters. $\triangle ABC \cong \triangle PQR$ pairs A with P, B with Q and C with R, in that order. So $AB = PQ$, $BC = QR$, $CA = RP$, $\angle A = \angle P$, and so on. Every other matching part then follows by CPCT (corresponding parts of congruent triangles are equal).

Question 5

If $\triangle PQR \cong \triangle ZXY$, which side is equal to QR, and which angle is equal to $\angle R$?

Answer. Pair the letters in order: P with Z, Q with X, R with Y. So QR = XY and $\angle R$ = $\angle Y$.

Time check: about 5 seconds.

Why SSA fails. In the figure, AB, angle B and the length of the side from A are all fixed, yet that side can end at $C_1$ or at $C_2$. Two different triangles fit the same data.

When it does not work:

• SSA and AAA prove nothing, as above.
• RHS is the one side-side-angle pattern that works, because the angle is $90^\circ$ and the side opposite it is the hypotenuse.
• Equal area or equal perimeter does not make two triangles congruent.

When SSC lists four conditions and asks which one is not enough, look for an angle that is not between the two given sides, or for three angles with no side.

Triplets and the 30-60-90 and 45-45-90 Triangles

When to use it: a right triangle with two sides given, or one side and a $30^\circ$, $45^\circ$ or $60^\circ$ angle.

Fast way: before squaring anything, look for a known pattern.

• Triplets: 3, 4, 5; 5, 12, 13; 8, 15, 17; 7, 24, 25; 20, 21, 29; 9, 40, 41; 12, 35, 37; 11, 60, 61; 28, 45, 53; 33, 56, 65; 16, 63, 65; 13, 84, 85. Any multiple works too: divide the given sides by their HCF first, so 39 and 52 become 3 and 4, and the hypotenuse is $13 \times 5 = 65$.
• Making a triplet: for an odd leg $m$, the others are $\frac{m^2 - 1}{2}$ and $\frac{m^2 + 1}{2}$ (11 gives 60 and 61). For an even leg $m$, they are $\left(\frac{m}{2}\right)^2 - 1$ and $\left(\frac{m}{2}\right)^2 + 1$ (16 gives 63 and 65).
• 30-60-90: the sides opposite $30^\circ$, $60^\circ$ and $90^\circ$ are in the ratio $1 : \sqrt{3} : 2$.
• 45-45-90: the sides are in the ratio $1 : 1 : \sqrt{2}$.

Acute, right or obtuse. With $c$ the longest side: $a^2 + b^2 > c^2$ means acute, $= c^2$ right, $< c^2$ obtuse. Sides 7, 8 and 11 give $49 + 64 = 113 < 121$, so that triangle is obtuse. Sides 36, 77 and 85 give $1296 + 5929 = 7225 = 85^2$, a right triangle.

Question 6

In triangle ABC, $\angle B = 90^\circ$, $\angle C = 30^\circ$ and $AC = 14$ cm. Find AB and BC.

Answer. AC is the hypotenuse. AB faces the $30^\circ$ angle, so it is half the hypotenuse: AB = 7 cm. BC faces the $60^\circ$ angle, so BC = $7\sqrt{3}$ cm.

Time check: about 10 seconds.

When it does not work:

• A triplet is a shortcut only when the triangle is known to be right-angled. Sides 5, 12 and 14 are not a triplet.
• The hypotenuse is always the longest side. If the two given sides are a leg and the hypotenuse, subtract the squares; don't add them.
• In a 30-60-90 triangle the shortest side faces $30^\circ$. Swapping the $\sqrt{3}$ side and the short side is the usual trap.

The Centre-Angle Table

When to use it: a centre of triangle ABC is named (incentre, excentre, circumcentre or orthocentre), and angle BXC at that centre is asked from angle A, or the other way round.

Fast way: one table.

So $\angle BIC + \angle BEC = 180^\circ$. Also, $\angle BIC$ is always more than $90^\circ$ and $\angle BEC$ is always less than $90^\circ$.

Question 7

In an acute triangle ABC, O is the circumcentre and $\angle BOC = 104^\circ$. Find $\angle BIC$ and $\angle BHC$, where I is the incentre and H the orthocentre.

Answer. $2A = 104^\circ$, so $A = 52^\circ$. Then $\angle BIC = 90^\circ + 26^\circ$ = 116° and $\angle BHC = 180^\circ - 52^\circ$ = 128°.

Time check: about 15 seconds.

When it does not work:

• Circumcentre, obtuse A. O moves to the other side of BC, as in the figure below, and $\angle BOC = 360^\circ - 2A$. For $A = 125^\circ$ that is $110^\circ$, not $250^\circ$. If $A = 90^\circ$, O is the midpoint of BC.
• Orthocentre, obtuse B or C. H lies outside, beyond the obtuse vertex, and then $\angle BHC = A$. If A itself is obtuse, $180^\circ - A$ still holds. If $A = 90^\circ$, H is the vertex A.
• The incentre and excentre rows always hold. The usual trap is to use $90^\circ - \frac{A}{2}$ for the incentre.

[SSC Tip] Where the centres sit: the incentre is always inside; the circumcentre and orthocentre are inside for an acute triangle, at the midpoint of the hypotenuse and at the right-angle vertex for a right triangle, and outside for an obtuse one.

Centroid and Medians, and the Median to the Hypotenuse

When to use it: medians, the centroid G, lengths like AG or GD, or a right triangle with the midpoint of its hypotenuse.

Fast way:

• G cuts every median in the ratio $2 : 1$ from the vertex: $AG = \frac{2}{3}AD$ and $GD = \frac{1}{3}AD$.
• The three medians cut the triangle into six small triangles of equal area, so triangle GBC is one third of triangle ABC.
• Median length (Apollonius): $AB^2 + AC^2 = 2(AD^2 + BD^2)$.
• Sum of the squares of the medians $= \frac{3}{4} \times$ (sum of the squares of the sides).
• If medians BE and CF meet at right angles, then $AB^2 + AC^2 = 5BC^2$.

Question 8

The medians BE and CF of triangle ABC cut each other at right angles. If $AB = 19$ cm and $AC = 22$ cm, find BC.

Answer. $5BC^2 = 19^2 + 22^2 = 361 + 484 = 845$, so $BC^2 = 169$ and BC = 13 cm.

Time check: about 15 seconds.

The median to the hypotenuse

In a right triangle, the midpoint M of the hypotenuse is the circumcentre, so $MA = MB = MC$ = half the hypotenuse. Triangles ABM and ACM are then isosceles, so the angles at the base copy across, and the angle at M is double.

The converse is just as useful: if a median is half the side it meets, the angle at that vertex is $90^\circ$.

Question 9

In triangle PQR, $\angle Q = 90^\circ$ and M is the midpoint of PR. If $\angle QPR = 34^\circ$, find $\angle QMR$.

Answer. $MQ = MP$, so $\angle MQP = \angle P = 34^\circ$. Angle QMR is the exterior angle of triangle PMQ, so $\angle QMR = 34^\circ + 34^\circ$ = 68°.

Time check: about 10 seconds.

When it does not work:

• $AG = \frac{2}{3}$ of the median, not of a side.
• Only the median to the hypotenuse is half of anything. The medians to the legs follow Apollonius.
• The $5BC^2$ rule needs the two medians from B and C, and BC is the side between those two vertices.

Radii of Equilateral and Right Triangles

When to use it: the inradius $r$ or the circumradius $R$ of an equilateral triangle or a right triangle is asked.

Fast way:

• Equilateral, side $a$. The height is $h = \frac{\sqrt{3}}{2}a$. All four centres are the same point, which sits two thirds of the way down each height. So $R = \frac{2}{3}h = \frac{a}{\sqrt{3}}$, $r = \frac{1}{3}h = \frac{a}{2\sqrt{3}}$, and $R = 2r$. In short, $r : R : h = 1 : 2 : 3$.
• Right triangle, legs $a$ and $b$, hypotenuse $c$. $R = \frac{c}{2}$, because the midpoint of the hypotenuse is the circumcentre, and $r = \frac{a + b - c}{2}$.
• Any triangle. $r = \frac{\Delta}{s}$ and $R = \frac{abc}{4\Delta}$, where $\Delta$ is the area and $s$ is half the perimeter. (Area formulas are in Chapter 13.)

For an equilateral triangle of side 18 cm, $h = 9\sqrt{3}$, $R = 6\sqrt{3}$ and $r = 3\sqrt{3}$.

Question 10

The two shorter sides of a right triangle are 20 cm and 21 cm. Find its inradius and circumradius.

Answer. 20, 21, 29 is a triplet, so the hypotenuse is 29 cm. Then $r = \frac{20 + 21 - 29}{2}$ = 6 cm and $R = \frac{29}{2}$ = 14.5 cm.

Time check: about 10 seconds.

When it does not work:

• $r = \frac{a + b - c}{2}$ and $R = \frac{c}{2}$ work only with a right angle. For other triangles use $\frac{\Delta}{s}$ and $\frac{abc}{4\Delta}$.
• In the equilateral formulas, don't put the height in place of the side. If the height is given, $R = \frac{2}{3}h$ and $r = \frac{1}{3}h$ straight away.

For an equilateral triangle, the circumcircle's area is 4 times the incircle's, because $R = 2r$.

Incircle Tangent Lengths: $s - a$

When to use it: a circle inside a triangle (or a quadrilateral) touches all its sides, and a piece of a side is asked.

Fast way: the two tangents from a point to a circle are equal, so each vertex has one tangent length. With $s = \frac{a + b + c}{2}$, the tangent length from A is $s - a$, from B is $s - b$, and from C is $s - c$, where $a = BC$, $b = CA$ and $c = AB$.

• If the incircle touches BC at D, then $BD = s - b$ and $DC = s - c$.
• For a right triangle, the tangent length from the right-angle vertex equals $r$.
• Quadrilateral round a circle: $AB + CD = AD + BC$. Opposite sides add up to the same total.

Question 11

In triangle ABC, $AB = 11$ cm, $BC = 16$ cm and $CA = 9$ cm. The incircle touches BC at D. Find BD.

Answer. $s = \frac{11 + 16 + 9}{2} = 18$. BD is the tangent length from B, which is $s - CA = 18 - 9$ = 9 cm. (Then $DC = 18 - 11 = 7$ cm, and $9 + 7 = 16$ checks.)

Time check: about 15 seconds.

For a quadrilateral ABCD drawn round a circle with $AB = 7$, $BC = 12$ and $CD = 15$, we get $AD = 7 + 15 - 12 = 10$.

When it does not work:

• BD is $s$ minus the side opposite B, which is CA, not BC. Check by adding: $BD + DC$ must give BC.
• $AB + CD = AD + BC$ pairs opposite sides. Adjacent sides need not add up to anything.

BPT and the Area Ratio in One Line

When to use it: a line parallel to one side of a triangle cuts the other two sides, and a length or an area is asked.

Fast way: if $DE \parallel BC$ and $AD : DB = m : n$, then triangle ADE is similar to triangle ABC with scale $\frac{m}{m + n}$.

$$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} = \frac{m}{m + n}$$

$$\text{area } ADE : \text{area } ABC = m^2 : (m + n)^2$$

$$\text{area } ADE : \text{trapezium } DBCE = m^2 : \left[(m + n)^2 - m^2\right]$$

The first row is the midpoint theorem: the line joining two midpoints is parallel to the third side and half of it.

Question 12

In triangle ABC, $DE \parallel BC$ with D on AB and E on AC, and $AD : DB = 4 : 3$. If the area of trapezium DBCE is 99 sq cm, find the area of triangle ADE. Also find BC if $DE = 12$ cm.

Answer. Area ADE : DBCE $= 16 : (49 - 16) = 16 : 33$, so area ADE $= \frac{16}{33} \times 99$ = 48 sq cm. And $\frac{DE}{BC} = \frac{4}{7}$, so BC $= 12 \times \frac{7}{4}$ = 21 cm.

Time check: about 20 seconds.

When it does not work:

• $\frac{DE}{BC}$ is $\frac{AD}{AB}$, not $\frac{AD}{DB}$. With $4 : 3$, the scale is $\frac{4}{7}$, not $\frac{4}{3}$.
• Areas go by the square of the scale. Lengths, perimeters, medians and heights go by the scale itself.
• No parallel line, no ratio. If DE is not parallel to BC, this method does not apply.

A line parallel to BC that cuts the triangle into two equal areas has $\frac{AD}{AB} = \frac{1}{\sqrt{2}}$.

Angle Bisector: Splitting the Side, and the Angle with the Altitude

When to use it: the bisector of angle A meets BC at D, and BD, DC or a side is asked. Or both the bisector and the altitude from A are drawn and the angle between them is asked.

Fast way:

• Side split: $\frac{BD}{DC} = \frac{AB}{AC}$. So $BD = BC \times \frac{AB}{AB + AC}$ and $DC = BC \times \frac{AC}{AB + AC}$.
• Angle between the bisector AD and the altitude AL from the same vertex: $\angle DAL = \frac{|B - C|}{2}$.
• [Tier 2] The bisector of the exterior angle at A cuts BC extended at a point P with $\frac{PB}{PC} = \frac{AB}{AC}$.
• [Tier 2] Length of the bisector: $AD^2 = AB \times AC - BD \times DC$.

With $\angle B = 72^\circ$ and $\angle C = 38^\circ$, the angle between the bisector and the altitude from A is $\frac{72^\circ - 38^\circ}{2} = 17^\circ$.

Question 13

In triangle ABC, $AB = 16$ cm, $AC = 12$ cm and $BC = 21$ cm. The bisector of $\angle A$ meets BC at D. Find BD.

Answer. BD : DC $= 16 : 12 = 4 : 3$, so BD $= \frac{4}{7} \times 21$ = 12 cm (and DC = 9 cm).

Time check: about 10 seconds.

When it does not work:

• The ratio uses the two sides that make the angle, AB and AC. BC is the side being cut.
• $\frac{|B - C|}{2}$ is the angle between the bisector and the altitude from the same vertex A. It does not give the angle between the bisector and a median.

The Altitude to the Hypotenuse

When to use it: $\angle A = 90^\circ$ and AD is drawn perpendicular to the hypotenuse BC.

Fast way: the altitude splits the triangle into two smaller triangles, both similar to the big one. That gives five one-liners.

• $AD^2 = BD \times DC$
• $AB^2 = BD \times BC$ and $AC^2 = DC \times BC$
• $AD = \frac{AB \times AC}{BC}$
• $\frac{1}{AD^2} = \frac{1}{AB^2} + \frac{1}{AC^2}$
• $\frac{AB^2}{AC^2} = \frac{BD}{DC}$

Question 14

In triangle ABC, $\angle A = 90^\circ$ and $AD \perp BC$. If $BD = 3$ cm and $DC = 12$ cm, find AD, AB and AC.

Answer. $AD^2 = 3 \times 12 = 36$, so AD = 6 cm. BC = 15, so $AB^2 = 3 \times 15 = 45$ and AB = $3\sqrt{5}$ cm, and $AC^2 = 12 \times 15 = 180$, so AC = $6\sqrt{5}$ cm.

Time check: about 20 seconds.

With the legs given instead, say 7 and 24, the hypotenuse is 25 and $AD = \frac{7 \times 24}{25} = \frac{168}{25}$ cm.

When it does not work:

• $AB^2 = BD \times BC$ uses the whole hypotenuse BC, not DC. Each leg goes with the piece next to it.
• The results need the right angle at A and the perpendicular dropped onto the hypotenuse. An altitude onto a leg gives nothing like this.

Chord Distance with Triplets

When to use it: of the radius, a chord's length and its distance from the centre, two are known and the third is asked. Also two parallel chords.

Fast way: the perpendicular from the centre cuts the chord in half, so half the chord, the distance and the radius make a right triangle.

$$d^2 + \left(\frac{l}{2}\right)^2 = r^2$$

Spot the triplet before squaring: radius 13 with chord 24 (half 12) gives distance 5.

• Equal chords are equally far from the centre, and the nearer chord is the longer one.
• Two parallel chords on the same side of the centre are $d_1 - d_2$ apart; on opposite sides, $d_1 + d_2$ apart.

Question 15

In a circle of radius 53 cm, AB and CD are parallel chords of lengths 90 cm and 56 cm. Find the distance between them if they are (i) on opposite sides of the centre, (ii) on the same side.

Answer. For AB, half is 45, so its distance is 28 (the 28, 45, 53 triplet). For CD, half is 28, so its distance is 45. Opposite sides: $28 + 45$ = 73 cm. Same side: $45 - 28$ = 17 cm.

Time check: about 15 seconds.

When it does not work:

• The right triangle uses half the chord. Putting in the full chord is the usual slip.
• A chord cannot be longer than the diameter. If your numbers make $\frac{l}{2}$ bigger than $r$, recheck the data.
• If a question with parallel chords does not say which side, both answers are possible. Look at the options to see which one is listed.

[SSC Tip] For two circles that cross, the common chord is perpendicular to the line joining the centres and is cut in half by it. When $r_1^2 + r_2^2 = d^2$, as with radii 15 and 20 and centres 25 apart, half the common chord is $\frac{r_1 r_2}{d} = 12$, so the chord is 24.

Angles in a Circle in One Look

When to use it: points on a circle, often with the centre O, and an angle is asked.

Fast way: first find the arc (or chord) the angle stands on. Then one of these finishes it.

• The angle at the centre is twice the angle at the circumference on the same arc.
• Angles on the same arc, in the same segment, are equal.
• The angle in a semicircle is $90^\circ$.
• An angle from the other side of the chord (the minor arc) is $180^\circ$ minus the angle from the major arc. Together they make a cyclic quadrilateral.
• In a cyclic quadrilateral, opposite angles add to $180^\circ$, and each exterior angle equals the interior opposite angle.
• Triangle OAB is isosceles ($OA = OB$), so $\angle OAB = 90^\circ - \angle ACB$ when C is on the major arc.

Question 16

A, B, C and D lie on a circle with centre O, with C on the major arc AB and D on the minor arc AB. If $\angle OAB = 28^\circ$, find $\angle ACB$ and $\angle ADB$.

Answer. $OA = OB$, so $\angle OBA = 28^\circ$ and $\angle AOB = 180^\circ - 56^\circ = 124^\circ$. Then $\angle ACB = \frac{124^\circ}{2}$ = 62°, and ACBD is cyclic, so $\angle ADB = 180^\circ - 62^\circ$ = 118°.

Time check: about 15 seconds.

When it does not work:

• The doubling rule needs both angles on the same arc. If C is on the minor arc, $\angle ACB = 180^\circ - \frac{1}{2}\angle AOB$.
• A quadrilateral is cyclic only if all four corners lie on the circle. A quadrilateral with O as a corner, like OACB, is not cyclic in general.
• $\angle OAB = 90^\circ - \angle ACB$ flips to $\angle ACB - 90^\circ$ when the angle at C is obtuse.

Tangent Length and the $180^\circ$ Rule

When to use it: tangents from an outside point P to a circle with centre O.

Fast way:

• The radius to the point of contact is perpendicular to the tangent: $OA \perp PA$.
• Tangent length: $PA = \sqrt{OP^2 - r^2}$. Triplets again: $OP = 37$ and $r = 12$ give $PA = 35$.
• The two tangents from P are equal, $PA = PB$, and OP bisects both $\angle APB$ and $\angle AOB$.
• $\angle APB + \angle AOB = 180^\circ$, because OAPB has two right angles.
• The chord AB makes equal angles with the tangents: $\angle PAB = \angle PBA = 90^\circ - \frac{P}{2}$. This is also the angle ACB at any point C on the major arc (alternate segment theorem).

Question 17

PA and PB are tangents to a circle with centre O, and $\angle APB = 52^\circ$. Find $\angle AOB$ and $\angle ACB$, where C is a point on the major arc AB.

Answer. $\angle AOB = 180^\circ - 52^\circ$ = 128°. The angle at the circumference is half of that, so $\angle ACB$ = 64°. (And $\angle PAB = 90^\circ - 26^\circ = 64^\circ$ too.)

Time check: about 10 seconds.

When it does not work:

• Use the distance from the centre. If P is 8 cm from the nearest point of a circle of radius 5 cm, then $OP = 8 + 5 = 13$ and the tangent is $\sqrt{169 - 25} = 12$ cm, not $\sqrt{64 - 25}$.
• The $180^\circ$ rule is for two tangents from the same point. For a tangent and a chord, use the alternate segment theorem: the angle between them equals the angle in the other segment.
• For C on the minor arc, $\angle ACB = 180^\circ - \frac{1}{2}\angle AOB$.

Tangent-Secant and Crossing Chords

When to use it: a line from an outside point P cuts the circle at A and B, and there is a tangent PT or another line from P. Or two chords cross inside the circle.

Fast way: use the power of a point.

• Tangent and secant from P: $PT^2 = PA \times PB$, with PA the near part and PB the whole length to the far point.
• Two secants from P: $PA \times PB = PC \times PD$.
• Two chords crossing at E inside: $AE \times EB = CE \times ED$.
• [Tier 2] Angles from arcs: two chords meeting inside make an angle equal to half the sum of the two arcs they cut off; two secants (or tangents) meeting outside make an angle equal to half the difference of the arcs.

Question 18

From an outside point P, a secant meets a circle at A and B, with $PA = 4$ cm and $AB = 12$ cm. Find the tangent PT. Also, chords AB and CD of another circle cross at E with $AE = 6$, $EB = 4$ and $CE = 3$. Find ED.

Answer. $PB = 4 + 12 = 16$, so $PT^2 = 4 \times 16 = 64$ and PT = 8 cm. For the chords, $6 \times 4 = 3 \times ED$, so ED = 8.

Time check: about 15 seconds.

When it does not work:

• PB is the whole length from P to the far point. Using $PA \times AB = 48$ is the trap.
• For chords crossing inside, multiply the two pieces of the same chord. Don't mix a piece of one chord with a piece of the other on the same side.
• If the outside angle is asked with arcs, take half the difference; half the sum is only for chords crossing inside.

Direct or Transverse Common Tangent

When to use it: two circles with radii $R$ and $r$ and centres $d$ apart, and the length of a common tangent between its two points of contact is asked.

Fast way:

• Direct tangent (both circles on the same side of it): $\sqrt{d^2 - (R - r)^2}$.
• Transverse tangent (it passes between the circles): $\sqrt{d^2 - (R + r)^2}$.

Memory hook: direct goes with the difference, transverse with the total. The transverse one is always the shorter.

• Circles touching externally: $d = R + r$, so the transverse length is 0 and the direct one is $\sqrt{(R + r)^2 - (R - r)^2} = 2\sqrt{Rr}$.
• A transverse tangent exists only when the circles are clear of each other, $d > R + r$.

Question 19

Two circles of radii 17 cm and 7 cm have their centres 26 cm apart. Find the lengths of the direct and the transverse common tangents.

Answer. Direct: $\sqrt{26^2 - 10^2} = \sqrt{576}$ = 24 cm. Transverse: $\sqrt{26^2 - 24^2} = \sqrt{100}$ = 10 cm.

Time check: about 15 seconds.

When it does not work:

• Mixing the two formulas is the standard trap, and the setters put both answers in the options. Here 24 and 10 would both be there.
• $d$ is the distance between the centres, not the gap between the circles. If the gap is given, add both radii to it.
• If a radius is the unknown, use the same formula backwards: a direct tangent of 24 with centres 26 apart gives $R - r = 10$.

Touching Circles and Counting Common Tangents

When to use it: two or three circles touch each other, or the number of common tangents is asked.

Fast way:

• Touching externally: the distance between the centres is $R + r$. Touching internally: $R - r$.
• Number of common tangents:

• Three circles touching each other externally: the centres make a triangle with sides $r_1 + r_2$, $r_2 + r_3$ and $r_3 + r_1$. Its perimeter is $2(r_1 + r_2 + r_3)$, and each radius is half that perimeter minus the side across from that circle's centre, the same $s - a$ idea as for incircle tangents.

Question 20

Three circles touch each other externally. The distances between their centres P, Q and R are $PQ = 11$ cm, $QR = 14$ cm and $PR = 17$ cm. Find the three radii.

Answer. Half the perimeter is $\frac{11 + 14 + 17}{2} = 21$. The radius at P is $21 - QR = 7$, at Q it is $21 - PR = 4$, and at R it is $21 - PQ = 10$. So the radii are 7 cm, 4 cm and 10 cm. Check: $7 + 4 = 11$, $4 + 10 = 14$, $7 + 10 = 17$.

Time check: about 20 seconds.

When it does not work:

• Internal touching uses the difference of the radii. A small circle touching a big one from inside has its centre $R - r$ from the big centre.
• Concentric circles (same centre) have no common tangent at all.

A Rough Figure, and Elimination by Size

When to use it: every question in words, and every time the options are far apart.

Fast way: spend five seconds on a rough figure that is roughly to scale. Draw a big angle big, keep equal sides equal, and mark every given length and angle. Then run these size checks on the options before any working.

• Triangle inequality: each side is less than the sum of the other two and more than their difference.
• Biggest side, biggest angle: the longest side faces the largest angle. The hypotenuse is the longest side of a right triangle.
• Centres: $\angle BIC$ is always above $90^\circ$; $\angle BEC$ at an excentre is always below $90^\circ$; in an equilateral triangle $r < R$, with $R = 2r$.
• Circles: a chord is at most the diameter; a tangent from P is shorter than OP; the transverse common tangent is shorter than the direct one.
• Right triangles: the altitude to the hypotenuse is shorter than both legs, and the median to the hypotenuse is exactly half of it.

Question 21

In triangle ABC, $AB = 7$ cm and $AC = 12$ cm. Which of these can be the length of BC: (a) 4 cm (b) 5 cm (c) 18 cm (d) 20 cm? Also, how many whole-number lengths are possible for BC?

Answer. BC must be more than $12 - 7 = 5$ and less than $12 + 7 = 19$. So (a), (b) and (d) are out, and the answer is (c) 18 cm. The whole numbers from 6 to 18 work, which is 13 lengths.

Time check: about 10 seconds.

When it does not work:

• If two options pass every check, work the question out in full.
• A printed figure may not be to scale, and SSC figures often are not. Trust the given numbers, never a measurement off the paper.
• The triangle inequality is strict: with sides 5, 7 and 12 the three points lie on one line, so no triangle is formed.

[SSC Tip] Trap options here follow a short list: $90^\circ - \frac{A}{2}$ in place of $90^\circ + \frac{A}{2}$, $2A$ for an obtuse angle at the circumcentre, $R + r$ in the direct tangent, the full chord in place of half, $\frac{AD}{DB}$ in place of $\frac{AD}{AB}$, and an area ratio that was not squared. If your answer equals one of these, go back and check that step.