⭕ Properties of Circles — Revision Notes

Walkthrough 1 — a chord and its distance from the centre Basic

1. AM = MB (⊥ bisector of chord) $=\frac{16}{2}=8$ cm
   OM starts at the centre and is perpendicular to the chord — two of the three conditions, so the third follows and OM bisects AB. Naming the property is what this line earns its mark for.
2. Join OA. Then OA is a radius, and ∠AMO = 90°.
   The question asks for the radius, so a radius has to appear in the picture before it can appear in the algebra. Joining OA closes triangle OAM, and the right angle at M was given.
3. OA² = AM² + OM² (Pythagoras' Theorem)
   = 8² + 15² = 64 + 225 = 289
   Triangle OAM is right-angled at M, so the radius OA is the hypotenuse. Half the chord and the distance from the centre are the two shorter sides — that is the shape every 8.1 length question takes.
4. OA = $\sqrt{289}$ = 17 (since OA > 0)
   ∴ the radius is 17 cm.
   Take only the positive root: a length cannot be negative. The answer is exact, so it is not rounded, and it is written as a sentence naming what was asked for — the radius, not just "OA".

Walkthrough 2 — the same triangle, run backwards Basic

1. PM = MQ (⊥ bisector of chord)
   Same trigger as before: a line from the centre, perpendicular to a chord. Write the reason even though nothing is calculated yet — it is what licences the doubling at the end.
2. PM² = OP² − OM² (Pythagoras' Theorem)
   = 25² − 7² = 625 − 49 = 576
   Here the hypotenuse is known and a shorter side is wanted, so Pythagoras is used as a subtraction. OP is a radius, so OP = 25 — that substitution is what the line turns on.
3. PM = $\sqrt{576}$ = 24 cm
   Positive root again. What has been found is PM, half the chord, since M is its midpoint.
4. PQ = 2 × 24 = 48 cm
   Double it, because M is the midpoint. The question asked for the chord, and the chord is twice the half-chord you just found.

Walkthrough 3 — when the radius is the unknown inside the equation Intermediate

1. The perpendicular bisector of XY passes through the centre O (⊥ bisector of chord), so O lies on NP.
   This is version (iii) of the property, and it is the only way into the question: nothing is given about O at all, so the centre has to be placed before anything can be measured from it.
2. Let the radius be r cm. Then OP = r and ON = OP − NP = r − 4.
   Both OP and OX are radii, so one letter r does two jobs. N lies between O and P in the figure, which is why the 4 cm is subtracted rather than added.
3. XN = $\frac{24}{2}=12$ cm (⊥ bisector of chord)
   The same halving as before. N is on the perpendicular bisector, so it is the midpoint of XY — the word "bisector" in the question has already said so.
4. OX² = XN² + ON² (Pythagoras' Theorem)
   r² = 12² + (r − 4)²
   Triangle OXN is right-angled at N. Because r sits on both sides, this is now an equation to solve rather than a value to evaluate — which is exactly why the question is a tier harder than Walkthrough 1.
5. r² = 144 + r² − 8r + 16
   8r = 160
   Expand (r − 4)² in full: it is r² − 8r + 16. The r² then cancels from both sides and what is left is linear.
6. r = 20
   ∴ the radius is 20 cm.
   Check it against the picture before moving on: if r = 20 then ON = 16, and 12² + 16² = 144 + 256 = 400 = 20². The triangle closes, so the answer is right.

Walkthrough 4 — carrying a distance across to the other chord Basic

1. ∠ONP = 90° (⊥ bisector of chord)
   N is the midpoint of PQ and ON passes through the centre — two conditions, so the third follows and ON is perpendicular to PQ. Only now is ON allowed to be called the distance of PQ from O.
2. ON = distance of PQ from O = distance of AB from O (equal chords)
   = OM = 20 cm
   The two chords are equal, so they are equidistant from the centre. This is the whole of part (i): no calculation, one property, one named reason — and the reason is the only thing on the line that can earn a mark.
3. PN² = OP² − ON² (Pythagoras' Theorem)
   = 29² − 20² = 841 − 400 = 441
   Triangle OPN is right-angled at N by step 1, with the radius OP as its hypotenuse. Part (i) has just handed you the height, so part (ii) is Walkthrough 2 all over again.
4. PN = $\sqrt{441}$ = 21 cm
   Positive root. PN is half of PQ, because N is the midpoint — the question said so in its first sentence.
5. PQ = 2 × 21 = 42 cm
   Double the half-chord. As a check, AB comes to 42 cm too, because the two chords were given equal.

Walkthrough 5 — two parallel chords, and why there are two answers Intermediate

1. Let the chords be AB = 10 cm and CD = 24 cm, and let M and N be the feet of the perpendiculars from O to AB and to CD. Then MN is the distance between the chords.
   The distance between two parallel lines is measured along a common perpendicular. OM and ON are both perpendicular to that pair, so M, O and N lie on one straight line and MN is a single length.
2. AM = 5 cm and CN = 12 cm (⊥ bisector of chord)
   OM² = 13² − 5² = 169 − 25 = 144, so OM = 12 cm
   Halve each chord, then use the radius as the hypotenuse. The short chord is the one that lies far from the centre: 12 cm out for a 10 cm chord.
3. ON² = 13² − 12² = 169 − 144 = 25, so ON = 5 cm
   The same triangle for the long chord, and the numbers swap over: the 24 cm chord sits only 5 cm from the centre. Long chords lie near the centre.
4. Case 1 — the chords are on opposite sides of O.
   MN = OM + ON = 12 + 5 = 17 cm

   With O between the two chords, the two distances add. Nothing in the question fixed which side each chord lies on, so this is only half the answer — which is exactly why it asked for "the two possible distances".
5. Case 2 — the chords are on the same side of O.
   MN = OM − ON = 12 − 5 = 7 cm

   Now N lies between O and M, so the distances subtract. Draw the second picture: which distance comes off which is decided by the figure, and the result is a positive length.
6. ∴ the two possible distances are 17 cm and 7 cm.
   State both distances and say which is which. The question asked for two, and the two come from the two positions the chords can take — either side of the centre, or the same side.

Walkthrough 6 — the gap is given and the radius is not Advanced

1. Let AB = 16 cm and CD = 30 cm, with M and N the feet of the perpendiculars from O. Let the radius be r cm and let ON = d cm. Then OM = d + 7.
   Two unknowns need two letters. CD is the longer chord, so it lies nearer the centre; the 7 cm gap is therefore added to d to reach M, which the figure confirms.
2. AM = 8 cm and CN = 15 cm (⊥ bisector of chord)
   Halve both chords. The two halves are the bases of the two right-angled triangles that share the same hypotenuse length r, and it is that sharing that will produce the equation.
3. From triangle OCN: r² = 15² + d²
   From triangle OAM: r² = 8² + (d + 7)²
   Both triangles have a radius as hypotenuse, so both expressions equal r². Writing the two of them down side by side is the whole method; everything after this is algebra.
4. 225 + d² = 64 + d² + 14d + 49
   225 = 113 + 14d
   14d = 112, so d = 8
   Set the two right-hand sides equal and expand (d + 7)² = d² + 14d + 49 in full. The d² terms cancel, leaving a linear equation — the same collapse that happened in Walkthrough 3.
5. r² = 15² + 8² = 225 + 64 = 289
   r = 17 ∴ the radius is 17 cm.
   Substitute back into either equation and use the other as the check: 8² + 15² = 289 as well, since d + 7 = 15. The answer is exact, so it is not rounded.

Walkthrough 7 — the right angle a tangent gives you Basic

1. ∠OTP = 90° (tangent ⊥ radius)
   OT is the radius drawn to the point of contact, so the tangent meets it square. Written first, this line turns the figure into a right-angled triangle OTP — and everything else follows from that.
2. OP² = OT² + PT² (Pythagoras' Theorem)
   = 8² + 15² = 64 + 225 = 289
   OP is opposite the right angle, so it is the hypotenuse and the two given lengths add. Note that OP is not a radius — it runs from the centre to a point outside the circle.
3. OP = $\sqrt{289}$ = 17 cm
   Exact, so no rounding. 8, 15, 17 is a Pythagorean triple worth recognising — as are 3, 4, 5 and 5, 12, 13 and 7, 24, 25, all of which turn up in this chapter.
4. tan ∠OPT = $\frac{OT}{PT}=\frac{8}{15}$
   ∠OPT = ${tan}^{\mathrm{-1}}$ $\frac{8}{15}$ = 28.07…° = 28.1° (to 1 d.p.)
   From P, OT is opposite and PT is adjacent, so the ratio is tangent. This answer is not exact, and the rule at the front of the paper is 1 decimal place for an angle in degrees — not 3 significant figures.

Walkthrough 8 — the kite, in three lines Basic

1. TA = TB (tangents from ext. pt.), so triangle TAB is isosceles.
   Part (i) is about the triangle formed by the two tangents and the chord of contact. The only thing that makes that triangle special is that its two tangent sides are equal — so say so, and name the property.
2. ∠TAB = ∠TBA (base ∠s of isos. △) $=\frac{180°-48°}{2}$ = 66°
   Angle sum of the triangle, then split what is left equally between the two base angles. The bracketed reason is doing two jobs here: it says why the two base angles are equal, which is what licences dividing by 2.
3. ∠OAT = ∠OBT = 90° (tangent ⊥ radius)
   Part (ii) is about the quadrilateral OATB, not the triangle. Its two tangent corners are right angles — that is the only information the figure holds about them.
4. ∠AOB = 360° − 90° − 90° − 48° (∠ sum of quadrilateral)
   = 132°
   The four angles of OATB add to 360°. Quote ∠ sum of quadrilateral as the reason rather than the supplementary shortcut: the shortcut is not one of the listed properties, and this version is one line longer.

Walkthrough 9 — bringing trigonometry into the kite Intermediate

1. ∠OBP = 90° (tangent ⊥ radius)
   ∠BOP = 180° − 90° − 32° (∠ sum of △) = 58°
   Triangle OBP has all three angles now: the right angle from the tangent, the 32° given, and the rest. Notice that ∠OPB is half the angle between the tangents, because OP is the mirror line.
2. ∠AOB = 2 × 58° (tangents from ext. pt.) = 116°
   OP bisects ∠AOB, so the angle just found is half of it. That is version (iii) of Property 4, and naming it is what turns the doubling into a step rather than a jump.
3. tan ∠OPB = $\frac{OB}{PB}$, so tan 32° = $\frac{6}{PB}$
   OB is a radius, so OB = OA = 6 cm. In the right-angled triangle OBP, OB is opposite the 32° and PB is adjacent to it — opposite over adjacent is tangent.
4. PB = $\frac{6}{tan32°}$ = 9.6020… = 9.60 cm (to 3 s.f.)
   Rearranged, not re-derived. Keep the unrounded 9.6020 on your calculator for part (iii) and round only what you write down — rounding now and reusing 9.60 later is how a final answer drifts in the last digit.
5. Area of OAPB = 2 × area of △OBP (tangents from ext. pt., so the two triangles are congruent) $=2\times \frac{1}{2}\times 9.6020\times 6$
   = 57.612 = 57.6 cm² (to 3 s.f.)
   The kite is two copies of the same right-angled triangle, whose legs are PB and OB. Use the unrounded 9.6020 here, give the answer to 3 s.f. because it is a length-based quantity, and write the units as cm².

Walkthrough 10 — a tangent, a line through the centre, and a chord Intermediate

1. ∠OAT = 90° (tangent ⊥ radius).
   Let the radius be r cm. Then OA = OB = r and OT = r + 18.
   B is on the circle and on OT, so OB is a radius and the 18 cm is the rest of the way out to T. One letter now stands for three lengths, which is what makes an equation possible.
2. OA² + TA² = OT² (Pythagoras' Theorem)
   r² + 24² = (r + 18)²
   Triangle OAT is right-angled at A, and OT is its hypotenuse. Both sides carry r², so — as in Walkthroughs 3 and 6 — the squared terms are about to cancel.
3. r² + 576 = r² + 36r + 324
   36r = 252, so r = 7
   ∴ the radius is 7 cm, and OT = 25 cm.
   Expand the bracket in full and cancel. Writing OT = 25 cm here is worth the line: part (ii) needs it, and 7, 24, 25 confirms the triangle is right-angled with the numbers you have.
4. cos ∠AOT = $\frac{OA}{OT}=\frac{7}{25}$, so ∠AOT = 73.739…°
   In right-angled triangle OAT, OA is adjacent to the angle at O and OT is the hypotenuse. This is ∠AOB as well, because B lies on OT.
5. OA = OB (radii), so ∠OAB = ∠OBA (base ∠s of isos. △) $=\frac{180°-73.739°}{2}=53.130\dots °$
   Triangle OAB has two radii for sides, so it is isosceles — a fact worth reaching for every time a chord and a centre appear together. Its apex angle is the 73.739° just found.
6. Let K be the foot of the perpendicular from O to AB. Then AK = KB (⊥ bisector of chord) and
   AK = OA cos ∠OAK = 7 cos 53.130° = 4.2
   ∴ AB = 2 × 4.2 = 8.4 cm
   Back to Property 1: drop the perpendicular from the centre and the chord splits in half, giving a right-angled triangle with a known angle and a known hypotenuse. The answer is exact — 8.4, not 8.40.

Walkthrough 11 — a shaded region between a tangent and an arc Advanced

1. ∠OAC = 90° (tangent ⊥ radius).
   Let the radius be r cm. Then OA = OB = r and OC = r + 16.
   Same opening as Walkthrough 10, because it is the same figure wearing different letters: a tangent, and a line from the centre out through the circle to the same external point.
2. r² + 24² = (r + 16)² (Pythagoras' Theorem)
   r² + 576 = r² + 32r + 256
   32r = 320, so the radius is 10 cm.
   The r² cancels and a linear equation is left. Check it: OC = 26, and 10² + 24² = 100 + 576 = 676 = 26². 10, 24, 26 is 5, 12, 13 doubled.
3. tan ∠AOC = $\frac{AC}{OA}=\frac{24}{10}$
   ∠AOC = 67.380…° = 67.4° (to 1 d.p.)
   From O, AC is opposite and OA is adjacent. Give the written answer to 1 d.p. as an angle, but keep 67.380° on the calculator — part (iii) divides by 360 and small errors magnify.
4. Area of △OAC = $\frac{1}{2}\times 10\times 24$ = 120 cm²
   The right angle at A makes OA and AC the base and the height, so no trigonometric area formula is needed. The shaded region is this triangle with the sector cut out of it.
5. Area of sector OAB = $\frac{67.380°}{360°}\times \pi \times {10}^2$ = 58.800… cm²
   The sector is the fraction $\frac{\theta }{360}$ of the whole circle, and ∠AOB is the same angle as ∠AOC because B lies on OC. ⓘ Mensuration
6. Shaded area = 120 − 58.800… = 61.199… = 61.2 cm² (to 3 s.f.)
   Triangle minus sector, in that order, because the sector lies inside the triangle. Round once, at the very end, to 3 significant figures, and give the units as cm².

Walkthrough 12 — halving at the circumference, and the isosceles triangle Basic

1. ∠APB = $\frac{126°}{2}$ (∠ at centre = 2 ∠ at ⊙^ce) = 63°
   Both angles stand on the same arc AB — the lower one, which P is not on — so the angle at the circumference is half the angle at the centre. The bracketed reason is the mark; the division is arithmetic.
2. OA = OB (radii), so triangle OAB is isosceles.
   Part (ii) is about the triangle at the centre, not about the arc. The only special thing about it is that two of its sides are radii — say so before using it.
3. ∠OAB = ∠OBA (base ∠s of isos. △) $=\frac{180°-126°}{2}=$ 27°
   The angle sum of the triangle leaves 54° to share between the two equal base angles. Quote both reasons — the isosceles one for why they are equal, and the angle sum for the subtraction.

Walkthrough 13 — when the angle at the centre is reflex Intermediate

1. ∠COD = 2 × ∠CAD (∠ at centre = 2 ∠ at ⊙^ce)
   = 2 × 26° = 52°
   Both angles stand on arc CD, the short one on the right that A is not on. The angle at the centre is the double, so this time you multiply rather than divide.
2. ∠AOC = 180° − 52° (adj. ∠s on a str. line) = 128°
   AOD is a straight line, so the angles AOC and COD along it add to 180°. This is the step the diameter was put there for.
3. Reflex ∠AOC = 360° − 128° (∠s at a point) = 232°
   B lies on the minor arc AC, so ∠ABC stands on the major arc — and the angle at the centre on that same arc is the reflex one. That is why the reflex angle is the one worked out here.
4. ∠ABC = $\frac{232°}{2}$ (∠ at centre = 2 ∠ at ⊙^ce) = 116°
   Half the reflex angle. Sanity-check it against the picture: ∠ABC looks obtuse there, and 116° is obtuse — 64° would have been visibly wrong.

Walkthrough 14 — a right angle handed to you free Basic

1. ∠ACB = 90° (rt. ∠ in semicircle)
   AB is a diameter and C is on the circle, so the angle at C is a right angle. Write this line before anything else: it is what turns the picture into a right-angled triangle, and without it neither part can be done.
2. AC² = AB² − BC² (Pythagoras' Theorem)
   = 26² − 10² = 676 − 100 = 576
   The right angle is at C, so the diameter AB is the hypotenuse — the longest side is the one through the centre. A leg is wanted, so this is a subtraction.
3. AC = $\sqrt{576}$ = 24 cm
   Exact, so no rounding. 10, 24, 26 is 5, 12, 13 doubled — recognising it is a useful check that no digit has slipped.
4. cos ∠ABC = $\frac{BC}{AB}=\frac{10}{26}$
   ∠ABC = 67.38…° = 67.4° (to 1 d.p.)
   At B, BC is adjacent and the diameter AB is the hypotenuse, so use cosine. Use the given lengths rather than the 24 you calculated: if part (i) had gone wrong, part (ii) would still be right.

Walkthrough 15 — a semicircle, a point outside, and a radius at the end Intermediate

1. ∠POB = 2 × ∠PAB (∠ at centre = 2 ∠ at ⊙^ce)
   = 2 × 24° = 48°
   ∠PAB at the circumference and ∠POB at the centre both stand on arc PB, the short arc at the bottom right that A is not on. Double it.
2. ∠APB = 90° (rt. ∠ in semicircle), and since PBQ is a straight line, ∠APQ = 90° too.
   AB is a diameter, so the angle at P is a right angle. Because Q is on PB produced, the ray PQ is the ray PB — so the same 90° serves triangle APQ.
3. In △APQ: ∠PAQ = 180° − 90° − 28° (∠ sum of △) = 62°
   ∴ ∠BAQ = 62° − 24° = 38°
   Work out the whole angle at A in the big triangle, then take away the part of it below AB. Subtracting rather than adding is settled by the figure: AB lies inside ∠PAQ.
4. In right-angled △APB: cos ∠PAB = $\frac{AP}{AB}$, so cos 24° = $\frac{15}{AB}$
   Back to the small triangle, where the right angle from step 2 lives. AP is adjacent to the 24° and the diameter AB is the hypotenuse, so cosine is the ratio that links them.
5. AB = $\frac{15}{cos24°}$ = 16.4195… cm
   Rearranged. Keep every decimal on the calculator — the very next line halves this number, and a value rounded here would move the final answer.
6. Radius = $\frac{16.4195\dots }{2}$ = 8.2097… = 8.21 cm (to 3 s.f.)
   AB is a diameter, so the radius is half of it — the step this question keeps for last. It is a length, not an angle, so it goes to 3 significant figures.

Walkthrough 16 — two transfers and an exterior angle Intermediate

1. ∠BDC = ∠BAC (∠s in same segment) = 38°
   Both angles stand on the chord BC, and A and D are both on the same side of it — the major arc. So the angle simply transfers across. Naming the chord to yourself is what makes the pairing safe.
2. ∠DBA = ∠DCA (∠s in same segment) = 27°
   The second transfer, on the other chord: ∠DBA and ∠DCA both stand on AD, with B and C on the same side of it. ∠DCA is just ∠ACD written the other way round — the same angle.
3. In △ABX: ∠XAB = ∠CAB = 38° and ∠XBA = ∠DBA = 27°.
   X lies on AC and on BD, so the angle at A in the small triangle is the whole of ∠BAC, and the angle at B is the whole of ∠DBA.
4. ∠AXB = 180° − 38° − 27° (∠ sum of △) = 115°
   Ordinary triangle work, once the two circle properties have supplied the other two angles. That is the shape of nearly every angle chase: circle properties to move angles around, then triangle and straight-line facts to finish.

Walkthrough 17 — the two pairs, kept apart Basic

1. All four of A, B, C and D lie on the circle, so ABCD is a cyclic quadrilateral and its opposite angles are angles in opposite segments.
   Say what the shape is before using what it does. Which pairs are "opposite" is decided by the lettering order ABCD, not by how the figure happens to be drawn: A pairs with C, and B pairs with D.
2. ∠ADC = 180° − ∠ABC (∠s in opp. segments)
   = 180° − 85° = 95°
   B and D are the opposite pair here, separated by the diagonal AC. Subtract from 180°, and quote the reason on the same line — that reason is the only part of this step worth a mark.
3. ∠DAB = 180° − ∠BCD (∠s in opp. segments)
   = 180° − 104° = 76°
   The other pair, A with C, separated by the diagonal BD. Check the whole shape at the end: 85 + 104 + 95 + 76 = 360, which is the angle sum of a quadrilateral — so nothing has gone wrong.

Walkthrough 18 — two sides produced to a point outside Intermediate

1. ∠ADC = 180° − ∠ABC (∠s in opp. segments)
   = 180° − 68° = 112°
   ABCD is a cyclic quadrilateral, and B and D are its opposite pair — separated by the chord AC. The two produced lines play no part yet; part (i) is inside the circle.
2. ∠CDE = 180° − 112° (adj. ∠s on a str. line) = 68°
   ADE is a straight line, so the angles at D on either side of it add to 180°. Notice the result: 68° again, the same as ∠ABC — that is the exterior-angle pattern appearing on its own, written out in the two safe steps.
3. In △CDE, the three angles are ∠CDE = 68°, ∠DCE = 55° and ∠CED.
   Everything is now outside the circle: a plain triangle with two known angles. Listing them before adding up keeps you from mixing ∠DCE up with ∠BCD, which is its supplement.
4. ∠CED = 180° − 68° − 55° (∠ sum of △) = 57°
   Ordinary triangle work to finish. A useful check: the exterior angle ∠ADC of triangle CDE should equal the two interior opposite angles, and 55° + 57° = 112° — it does.

Walkthrough 19 — a centre and a cyclic quadrilateral together Intermediate

1. The angle wanted, ∠OAD, is part of ∠DAB: ∠OAD = ∠DAB − ∠OAB.
   Work backwards first. OA lies inside the angle DAB, so the target splits into two pieces — and each piece has its own route. Planning the split before calculating is what stops this question turning into guesswork.
2. ∠DAB = 180° − ∠BCD (∠s in opp. segments)
   = 180° − 78° = 102°
   ABCD is a cyclic quadrilateral, so A and C — the opposite pair — are supplementary. The centre plays no part in this step at all.
3. OA = OB (radii), so ∠OAB = ∠OBA (base ∠s of isos. △) $=\frac{180°-96°}{2}$ = 42°
   The second piece. The 96° at the centre is the apex angle of the isosceles triangle OAB, so the two base angles share the remaining 84° equally.
4. ∠OAD = 102° − 42° = 60°
   Subtract, because OA lies between AB and AD in the figure. Two properties and one subtraction, with the two halves of the question meeting only on this line.

Walkthrough 20 — a tangent, a diameter and a cyclic quadrilateral in one figure Advanced

1. ∠ACD = 180° − ∠AED (∠s in opp. segments)
   = 180° − 118° = 62°
   A, E, D and C lie on the circle, so AEDC is a cyclic quadrilateral with E and C as its opposite pair, separated by the chord AD. Five points offer several cyclic quadrilaterals; this is the one holding the given 118°.
2. ∠DBA = ∠DCA (∠s in same segment) = 62°
   Both stand on the chord AD, with B and C on the same side of it. This is the transfer that carries the 62° from the top of the figure down to B, where the centre can reach it.
3. O lies on BD, so ∠OBA = ∠DBA = 62°.
   OA = OB (radii), so ∠OAB = ∠OBA (base ∠s of isos. △) = 62°
   BD is a diameter, so the centre is on it and the ray BO is the ray BD — that is what makes the transferred angle usable. Triangle OAB then has two radii for sides.
4. ∠OAR = 90° (tangent ⊥ radius)
   ∴ ∠BAR = 90° − 62° = 28°
   The radius OA goes to the point of contact, so it meets the tangent square, and AB lies inside that right angle. Two lines, both on the syllabus, give what the alternate segment theorem would have given in one.
5. In △XCD: ∠XDC = ∠BDC = 24° and ∠XCD = ∠ACD = 62°, so
   ∠DXC = 180° − 24° − 62° (∠ sum of △) = 94°
   X is on BD and on AC, so the angles of the little triangle at D and at C are exactly the two angles already known. The 24° has waited four steps for its moment.
6. ∠AXD = 180° − 94° (adj. ∠s on a str. line) = 86°
   AXC is a straight line, so ∠AXD and ∠DXC are supplementary. Twenty marks' worth of figure, six lines of working, and every line names the one property it used — which is the whole skill this chapter is testing.