Circles: Questions and Answers

70 circles questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• Can you recognise the origin of the shapes in the figure?Answer: The three circular shapes come from raindrops falling on water (ripples), the cross-section of a plant stem or tree trunk (growth rings) and the flower head (inflorescence) of a sunflower.
• What properties are common to all circles, big and small?Answer: Every circle, big or small, has a centre, and every point of the circle is the same distance from the centre (this distance is the radius). Only the length of the radius changes from one circle to another.
• List some objects from nature that resemble a circle.Answer: For example: the full moon, the disc of the sun, ripples on water, the cut face of a tree trunk, a sunflower head, the iris and pupil of an eye, a slice of orange or lemon, and the cap of a mushroom seen from above.
• Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?Answer: Amina most likely said: fold the circle in half so that its edges overlap exactly, open it, then fold it in half again along a different direction. The two creases are diameters, and they cross at the centre.
• Say you are looking at a wheel of a vehicle. You see a point of the wheel touching the ground. When you look at the wheel again after some time, you again see a point of the wheel touching the ground. Can you tell if the two points are the same point?Answer: No. A rotating circular wheel looks exactly the same in every position, so there is no way to tell whether the point touching the ground is the same point as before. The circle has complete rotational symmetry.
• Draw a circle on the paper and cut along the circle. Fold the circular paper so that the boundaries overlap, then open it. You see a crease; it is a line of reflection symmetry of the circle. Does this line pass through the centre of the circle?Answer: Yes. The crease always passes through the centre: it is a diameter. Every diameter of a circle is a line of reflection symmetry.
• 1. What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?
  2. What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?
  3. The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points?
  (Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)Answer: 1. Square: turns of 90°, 180°, 270° and 360° about its centre; 4 lines of symmetry. Regular pentagon: turns of 72°, 144°, 216°, 288°, 360°; 5 lines. Regular hexagon: turns of 60°, 120°, …, 360°; 6 lines. 2. The longest chord is a diameter, 10 units; there is no smallest chord. 3. The locus is the perpendicular bisector of the segment joining the two points.
• 1. How many circles pass through two points on a plane?
  2. Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?
  3. As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?
  4. As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?
  5. You are given two points A and B on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?Answer: 1. Infinitely many. 2. No: the radius must be at least ½AB. Smallest radius = ½AB (centre at the midpoint); there is no largest. 3. The radii increase. 4. The circles look less curved (flatter near AB). 5. Infinitely many squares have A and B on the boundary; exactly 3 squares have A and B as corners.
• How many circles can you draw through three distinct points A, B and C on a plane? Is there always at least one such circle? Not necessarily! What if A, B and C lie on a straight line, i.e., are collinear? Can you explain why, in this case, there is no circle through A, B and C?Answer: The centre of a circle through A, B and C would have to lie on the perpendicular bisector of AB and on the perpendicular bisector of BC. When A, B, C are collinear these two bisectors are both perpendicular to the same line, so they are parallel and never meet. No point can be the centre, so no circle passes through three collinear points.
• Let us assume that A, B and C are not collinear. Is there always a circle passing through A, B and C? Can there be more than one circle through A, B and C?Answer: Yes, always exactly one. There is a unique circle passing through three non-collinear points. Its centre is the point where the perpendicular bisectors of AB and AC meet.
• Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?Answer: ∠C = 50°, so all three angles are acute. The circumcentre lies inside the triangle. (The circumradius comes out about 3.3 cm.)
• Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?Answer: ∠A = 100° is obtuse, so the circumcentre lies outside the triangle (beyond the side BC, opposite A). The circumradius comes out about 3.5 cm.
• Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.Answer: OA = OB = OC ≈ 3.9 cm (exactly 3.87 cm to two decimal places). All three are equal: each is the radius of the circumcircle.
• What is the least possible radius of a circle through two points A and B?Answer: The least possible radius is ½AB (half the distance between A and B). This circle has its centre at the midpoint of AB, and AB is a diameter.
• 1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC ? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
  2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?Answer: 1. No such point P exists. The perpendicular bisectors of AB and BC are both perpendicular to the line ABC, so they are parallel. Hence no circle passes through three collinear points, and no line cuts a circle in three distinct points (at most two). 2. Yes, infinitely many: turn ΔABC about the centre through any angle, or reflect it in any diameter.
• Equal chords of a circle subtend equal angles at the centre of the circle.Answer: If chords AB and DE of a circle with centre C are equal, then ΔCAB ≅ ΔCDE by SSS (CA = CD, CB = CE as radii, AB = DE given), so ∠ACB = ∠DCE.
• Chords of a circle that subtend equal angles at the centre are equal.Answer: If ∠ACB = ∠DCE at the centre C, then ΔACB ≅ ΔDCE by SAS (AC = DC, BC = EC as radii, with the equal angles between them), so AB = ED.
• Show that the triangle formed by a chord and the centre of the circle is isosceles.Answer: For a chord AB of a circle with centre O, the two sides OA and OB of ΔOAB are both radii, so OA = OB and ΔOAB is isosceles (and so ∠OAB = ∠OBA).
• Show that if two such isosceles triangles (formed by a chord and the centre of the circle) have equal base length, they are congruent to each other.Answer: For equal chords AB and CD of a circle with centre O: OA = OC, OB = OD (radii) and AB = CD (equal bases), so ΔOAB ≅ ΔOCD by SSS.
• The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.Answer: With centre C, chord AB and midpoint M: ΔCMA ≅ ΔCMB (SAS: CA = CB, ∠A = ∠B, AM = BM), so ∠CMA = ∠CMB. They add up to 180°, so each is 90°: CM ⟂ AB.
• Can you explain why the converse of the result “the line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord” is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?
  (Hint: Use the figure. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)Answer: If CM ⟂ AB, then in right triangles CMA and CMB the hypotenuses CA = CB (radii) and CM is common, so ΔCMA ≅ ΔCMB (RHS) and AM = BM: the perpendicular bisects the chord.
• An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.Answer: Both A (since AB = AC) and the centre O (since OB = OC) are equidistant from B and C, so both lie on the perpendicular bisector of BC. That line passes through A and is perpendicular to BC, so it is the altitude from A; hence the altitude passes through O.
• Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.Answer: The 6 cm chord is 4 cm from the centre and the 8 cm chord is 3 cm from it. On opposite sides, the midpoints are 4 + 3 = 7 cm apart.
• Take a paper circle. Fold the circle from the boundary, inwards. Open the fold. The crease is now a chord (see the figure B). Now fold the paper again, so that the end points of the chord meet. Open the fold (see the figure C).
  Measure the lengths of the parts into which the chord is divided. The chord gets bisected where the folds intersect. Measure the angle between the creases. The crease of the second fold is along the perpendicular from the centre to the chord. Measure the distance from the centre to the midpoint of the chord. It is the distance from the centre to the chord.
  Now draw another chord of the same length. How will you do this? We will let you figure this out yourself. Join the centre to the midpoint of the new chord and measure its length. Is it the same as distance from the centre to the first chord?Answer: The two parts of the chord are equal and the creases meet at 90°. To draw another chord of the same length, open the compasses to the chord’s length, put the point anywhere on the circle and cut the circle. The new chord’s distance from the centre is the same as the first chord’s: equal chords are equidistant from the centre.
• Chords of a circle having the same length are all at the same distance from the centre of the circle.Answer: For equal chords AB and FG with midpoints E and H, CE ⟂ AB and CH ⟂ FG. Then ΔCEA ≅ ΔCHF by RHS (CA = CF radii, AE = FH, right angles), so CE = CH: equal chords are equidistant from the centre.
• Use the Baudhāyana–Pythagoras theorem to show why chords of a circle having the same length are all at the same distance from the centre of the circle.Answer: For a chord of length l at distance d from the centre of a circle of radius r: d² = r² − (l/2)². The right side is the same for two chords of equal length, so their distances are equal: equal chords are equidistant from the centre.
• Consider the figure. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.Answer: Right triangles CEA and CHF have CA = CF (radii) and CE = CH (given), so ΔCEA ≅ ΔCHF (RHS) and AE = FH. The perpendiculars from the centre bisect the chords, so AB = 2AE = 2FH = GF.
• If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF. Solve this using the Baudhāyana–Pythagoras theorem.Answer: AE² = CA² − CE² = r² − CE² and FH² = r² − CH². As CE = CH, AE = FH, and doubling the half-chords gives AB = GF.
• You have two chords on a circle. One is longer than the other. Which chord is closer to the centre? Can you guess?
  Activity: Draw a circle. Draw chords of various lengths. Drop a perpendicular to each chord from the centre. Record the length of the chord and its distance from the centre in a table.
  What do you observe?Answer: The longer chord is closer to the centre. In a circle of radius 5 cm, for example: chord 10 cm → 0 cm, 9.6 cm → 1.4 cm, 8 cm → 3 cm, 6 cm → 4 cm, 2.8 cm → 4.8 cm. As the chord gets shorter, its distance from the centre grows.
• Let AB and DE be two chords of a circle with centre C. Suppose AB > DE. Then the distance from C to AB is less than the distance from C to DE.Answer: Drop perpendiculars CF to AB and CG to DE. Then CF² + AF² = CA² = CD² = CG² + GD². Since AB > DE, AF > GD, so CF² < CG² and CF < CG: the longer chord is nearer the centre.
• Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.Answer: Half-chord = √(7² − 6²) = √13 cm, so the chord = 2√13 cm ≈ 7.21 cm.
• Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r² − d²).Answer: The perpendicular OM from the centre bisects the chord AB, and ΔOMB is right-angled at M with hypotenuse OB = r and leg OM = d. So MB = √(r² − d²) and AB = 2MB = 2√(r² − d²).
• In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.Answer: No. Chord length is 2√(r² − d²), which is not proportional to the distance d. For example, with radius 10 cm, CD at 3 cm is about 19.1 cm, AB at 6 cm is 16 cm, and 19.1 ≠ 2 × 16. All we can say is that AB is shorter than CD.
• A circle with centre O is drawn, and A, B, C, D are points on the circle (see the figure). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.Answer: Arc AKB subtends about 100° at O (less than 180°), so it is a minor arc. Arc CLD subtends about 200° at O (more than 180°), so it is a major arc.
• Draw a circle and a chord AB. Fix an arc AKB formed by AB and a point K between A, B on the circle. Measure the angle subtended at the centre by arc AKB. Take three points P, Q, R on the circle outside arc AKB. Measure the angles subtended by arc AKB at points P, Q, R. What do you notice?
  Repeat this activity for a different arc AKB.Answer: The angles at P, Q and R are all equal, and each is half the angle at the centre. For example: centre 80° → P, Q, R each 40°; with a different arc, centre 130° → each 65°.
• The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.Answer: With centre C, arc AFB and a point D on the circle outside the arc: join DC and extend it to E. In the isosceles triangles DCA and DCB, each exterior angle at C is twice the base angle at D. Adding (or subtracting) the two parts gives ∠ACB = 2∠ADB.
• The angle subtended by a diameter at any point on the circle is 90°.Answer: If AB is a diameter and D is any other point on the circle, the arc AB not containing D subtends a straight angle ∠ACB = 180° at the centre C, so ∠ADB = ½ × 180° = 90°.
• In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?Answer: OA = OB = 12 cm, so the base angles of ΔOAB are (180° − 60°) ÷ 2 = 60° each. The triangle is equilateral, so AB = 12 cm.
• Let A and B be two points on a circle with centre O.Answer: (i) No: points on the same side of AB see AB at equal angles. (ii) Yes, unless AB is a diameter: points on opposite sides see AB at angles adding to 180°, so they can be equal only if both are 90°. (iii) Yes, if X and Y are on the same side of AB (then A, B, X, Y are concyclic); if they are on opposite sides, not in general.
• Find x in the figure.Answer: ∠ADC and ∠ABC are opposite angles of the cyclic quadrilateral ADCB, so x = 180° − 100° = 80°.
• If a line segment AB joining two points A, B subtends equal angles at two other points C, D that lie on the same side of AB, then the four points lie on a circle.Answer: Draw the circle through A, B, C. If D were outside it, AD would meet the circle at E with ∠AEB > ∠ADB; if D were inside, AD extended would meet it at E with ∠ADB > ∠AEB. Either way ∠ACB = ∠AEB contradicts ∠ACB = ∠ADB. So D lies on the circle: A, B, C, D are concyclic.
• The sum of two opposite angles of a cyclic quadrilateral is 180°.Answer: ∠BAD is half the angle that arc BCD subtends at the centre O, and ∠BCD is half the angle that arc BAD subtends at O. These two angles at O make a full turn, 360°, so ∠BAD + ∠BCD = ½ × 360° = 180°.
• A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = 100°, and ∠D = 70°. Can such a quadrilateral be drawn? Explain why or why not.Answer: Yes. The angles add up to 360°, and the opposite angles are supplementary: ∠A + ∠C = 80° + 100° = 180° and ∠B + ∠D = 110° + 70° = 180°. A quadrilateral whose opposite angles add up to 180° is cyclic.
• If two opposite angles of a quadrilateral add up to 180°, then the vertices of the quadrilateral lie on a circle, i.e., they are concyclic.Answer: Take the circle through A, B, D. If C were outside it, CD would meet the circle at E and ∠BED (an exterior angle of ΔBEC) would be greater than ∠BCD; if C were inside, DC produced would meet it at E and ∠BCD would be greater than ∠BED. But ∠BCD = ∠BED in both cases (each is 180° − ∠BAD). So C lies on the circle: ABCD is cyclic.
• In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?Answer: Half-chord = √(13² − 5²) = √144 = 12 cm, so the chord is 24 cm.
• An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?Answer: The angle at a point on the circle (outside the arc) is half the angle at the centre: ½ × 70° = 35°.
• The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.Answer: Radius = 13 cm and half-chord = 12 cm, so the distance = √(13² − 12²) = √25 = 5 cm.
• A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?Answer: Half-chord = √(15² − 9²) = √144 = 12 cm, so the chord is 24 cm.
• Prove that the perpendicular bisector of a chord passes through the centre of the circle.Answer: The centre O is equidistant from the ends of every chord AB (OA = OB, radii). Every point equidistant from A and B lies on the perpendicular bisector of AB, so O lies on it.
• The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.Answer: ∠ACB = 90°. The arc AB not containing C subtends a straight angle (180°) at the centre O, so it subtends ½ × 180° = 90° at C.
• ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?Answer: Opposite angles of a cyclic quadrilateral add up to 180°: ∠C = 180° − 75° = 105° and ∠D = 180° − 110° = 70°.
• Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find the value of x and the measures of ∠P and ∠R.Answer: ∠P + ∠R = 180°: 5x − 10 = 180, so x = 38, ∠P = 86° and ∠R = 94°.
• The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.Answer: Half-chord = 8 cm, so radius = √(8² + 6²) = √100 = 10 cm.
• A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.Answer: Whichever order the sides are in, two opposite angles turn out to be 90°, so the quadrilateral splits into two right triangles with legs 5 and 12: area = 2 × ½ × 5 × 12 = 60 square units.
• Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?Answer: Look at the angle each side subtends at an opposite vertex, for example side AB at C (∠ACB). If all four such angles are less than 90°, the centre is inside; if one is 90°, the centre is on that side; if one is more than 90°, the centre is outside, beyond that side. This needs only a protractor, so it is the best way.
• When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.Answer: Let equal chords AB and CD meet at P, with OM ⟂ AB and ON ⟂ CD. Equal chords are equidistant, so OM = ON; then ΔOMP ≅ ΔONP (RHS) gives MP = NP. Also AM = CN (half of equal chords). So AP = AM + MP = CN + NP = CP, and then PB = PD.
• Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.
  (Hint: Is it a circumcircle of a suitable triangle?)Answer: Draw right triangle ABD with AB = 6 cm, BD = 6 cm and ∠B = 90°. Its circumcircle has centre O at the midpoint of the hypotenuse AD, so O is 3 cm from AB. The required radius is 3√2 ≈ 4.24 cm, and AB is the required chord.
• Show that rectangle is the only parallelogram that can be inscribed in a circle.Answer: If parallelogram ABCD is cyclic, then ∠A = ∠C (parallelogram) and ∠A + ∠C = 180° (cyclic), so ∠A = ∠C = 90°; all angles are then 90° and ABCD is a rectangle. Conversely every rectangle is cyclic, since its opposite angles add up to 180°.
• Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.Answer: ∠ABC = 90°, so the arc AC not containing B subtends 2 × 90° = 180° at the centre O: A, O, C are collinear and AC is a diameter. Likewise BD is a diameter. Two diameters meet only at the centre, so the diagonals meet at O.
• Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?Answer: A circle with the same centre. If the circle has radius r and the chords have length l, every midpoint is at distance √(r² − (l/2)²) from the centre. (If l = 2r, the chords are diameters and the “circle” shrinks to the centre itself.)
• In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.Answer: Join OA, OB, OC. In ΔOAB and ΔOAC: OB = OC (radii), AB = AC (given), OA is common, so ΔOAB ≅ ΔOAC (SSS) and ∠OAB = ∠OAC. So AO bisects ∠BAC: the centre lies on the bisector.
• Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.Answer: Let the 24 cm chord be x cm from the centre; the 10 cm chord is then (x + 7) cm away. r² = 12² + x² = 5² + (x + 7)² gives x = 5, so r = 13 cm.
• A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.Answer: Each side subtends 360° ÷ 6 = 60° at the centre, so each triangle formed with the centre is equilateral: side = r. The distance of each side from the centre is √(r² − (r/2)²) = (√3/2)r ≈ 0.866r.
• A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.Answer: ∠MOP = ∠MNP. Both angles stand on the chord MP, and O and N lie on the same side of MP (on the same arc), so they are angles in the same segment. Because MN is a diameter, ∠MPN = ∠MON = 90° as well, so ∠MOP = ∠MNP = 90° − ∠PMN.
• Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).Answer: (The question’s ∠CDE is the exterior angle at D when E is on AD produced; below, E is taken on CD produced, so the same exterior angle is called ∠ADE.) With E on CD produced beyond D, ∠ADE + ∠ADC = 180° (linear pair) and ∠ABC + ∠ADC = 180° (opposite angles of a cyclic quadrilateral). So the exterior angle ∠ADE = ∠ABC, the interior opposite angle.
• “There is no chord of a circle that is longer than its diameter.” How do you justify this statement?Answer: For any chord AB of a circle with centre O and radius r, the triangle inequality gives AB ≤ OA + OB = 2r, with equality only when O lies on AB, that is, when AB is a diameter. So no chord is longer than a diameter.
• Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.Answer: A chord through A at distance d from O has length 2√(r² − d²), which is shortest when d is largest. The perpendicular OM from O to any chord through A satisfies OM ≤ OA (OA is the hypotenuse of ΔOMA), with equality only when M = A, that is, when the chord is perpendicular to OA.
• How would you use the following figure to justify the statement that the angle in a semicircle is 90°?Answer: The ticks show OA equals the two half-diameters, so the two smaller triangles are isosceles: the angle at A splits into a and b. The angles of the big triangle then add to a + b + (a + b) = 180°, so a + b = 90°: the angle at A is 90°.
• In a circle, two chords CC’ and DD’ are drawn perpendicular to a diameter AB. Prove that the segment MM’ joining the midpoints of the chords CD and C’ D’ is perpendicular to AB.Answer: The diameter AB is perpendicular to CC′ and DD′, so it bisects them: C′ and D′ are the mirror images of C and D in AB. Using congruent triangles (or this reflection), CD and C′D′ are mirror images, so their midpoints M and M′ are mirror images too, and MM′ ⟂ AB.
• How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?Answer: OA = OB = OC = OD, so each of the four triangles OAB, OBC, OCD, ODA is isosceles: ∠OBA = p, ∠OCB = q, ∠ODC = u, ∠OAD = v. Then ∠A = p + v, ∠B = p + q, ∠C = q + u, ∠D = u + v. The four angles total 2(p + q + u + v) = 360°, so ∠A + ∠C = ∠B + ∠D = p + q + u + v = 180°.

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