Book page 114–116 Updated on2026-09-08
Chord = 24 cm.
35°, at any point of the circle lying outside that arc.
Distance = 5 cm.
Chord = 24 cm.
Given: a circle with centre O and a chord AB. To show: the perpendicular bisector of AB passes through O.
The locus step, written out. Let M be the midpoint of AB and join OM. In ΔOMA and ΔOMB: OA = OB (radii), AM = BM (M is the midpoint), OM common. By SSS, ΔOMA ≅ ΔOMB, so ∠OMA = ∠OMB. These are angles on a straight line, so ∠OMA + ∠OMB = 180°, giving ∠OMA = ∠OMB = 90°. Hence OM is perpendicular to AB and bisects it — OM is the perpendicular bisector, and it passes through O.
∠ACB = 90° — the angle in a semicircle is a right angle.
A second proof, using isosceles triangles. Join OC. Then OA = OC = OB = r, so ΔOAC and ΔOBC are both isosceles.
∠C = 105° and ∠D = 70°.
x = 38, ∠P = 86° and ∠R = 94°.
Radius = 10 cm.
Area = 60 square units.
Let the quadrilateral be ABCD with AB = BC = 5 and CD = DA = 12 — the two equal pairs adjacent, so it is a kite. Join the diagonal BD.
If instead the sides alternate 5, 12, 5, 12: opposite sides are equal, so the quadrilateral is a parallelogram; a cyclic parallelogram is a rectangle (Q14), and the area is 5 × 12 = 60 square units again.
Best method: draw one diagonal and look at the two triangles it makes.
A diagonal, say AC, cuts the cyclic quadrilateral ABCD into ΔABC and ΔACD. Both triangles are inscribed in the same circle, so the circumcentre of the quadrilateral is the circumcentre of each of them. And we already know where a triangle's circumcentre sits:
| Triangle ABC or ACD | Circumcentre lies | Conclusion for ABCD |
|---|---|---|
| one of them is acute-angled | inside that triangle | inside the quadrilateral |
| one of them is right-angled | at the midpoint of its hypotenuse | on the diagonal AC (AC is a diameter) |
| both are obtuse-angled | outside both triangles | outside the quadrilateral |
An equivalent test using the sides. Each side of the quadrilateral cuts off an arc. The centre lies inside exactly when no side cuts off an arc bigger than a semicircle. Since an inscribed angle is half its arc, this becomes a test you can carry out with a protractor:
Given: chords AB and CD of a circle with centre O, with AB = CD, meeting at a point P inside the circle. To show: the two pieces of AB match the two pieces of CD.
Step 1 — equal chords are equidistant from the centre. Let M and N be the midpoints of AB and CD. By Theorem 6, OM = ON, and OM ⊥ AB, ON ⊥ CD.
Step 2 — P is equidistant from the two midpoints.
Step 3 — put the pieces together. Since M and N are midpoints, AM = MB = ½AB and CN = ND = ½CD, and AB = CD gives AM = CN.
The circle is forced: its radius must be 3√2 ≈ 4.24 cm.
Construction (direct).
Answering the hint. Yes — it is the circumcircle of a triangle, and a very recognisable one.
So an equivalent construction is: draw ΔABC with AB = 6 cm and ∠ACB = 45°, and draw its circumcircle. That circle automatically has the chord AB = 6 cm standing 3 cm from the centre.
Given: a parallelogram ABCD inscribed in a circle. To show: ABCD is a rectangle.
The converse holds too, so nothing is lost: every rectangle is cyclic. Its diagonals are equal and bisect each other, so their common midpoint is the same distance from all four vertices — that point is the centre of a circle through A, B, C, D.
Given: rectangle ABCD inscribed in a circle. To show: the diagonals meet at the centre.
Proof 1 — each diagonal is a diameter.
Proof 2 — using the properties of a rectangle. The diagonals of a rectangle are equal and bisect each other. Let them meet at P. Then
Another circle, with the same centre — a concentric circle of radius √(r² − ℓ²/4), where r is the radius of the given circle and ℓ the fixed chord length.
And every point of that circle is reached. Take any point M with OM = √(r² − ℓ²/4). Draw the chord through M perpendicular to OM. By Theorem 5 it is bisected at M, and its length is 2√(r² − OM²) = 2 × (ℓ/2) = ℓ. So M really is the midpoint of a chord of length ℓ. The locus is the whole concentric circle, not just part of it.
Given: circle with centre O; chords AB and AC with AB = AC. To show: AO bisects ∠BAC.
A second way to see it. AB = AC are equal chords, so they are equidistant from O (Theorem 6). A point inside an angle that is equidistant from both arms lies on the bisector of that angle. Hence O lies on the bisector of ∠BAC.
Radius = 13 cm.
Side = r, and the distance of each side from the centre = (√3/2) r.
∠MOP = ∠MNP — they are equal, being angles in the same segment standing on the chord MP.
What the diameter MN gives you as well. Since MN is a diameter, the angle it subtends at any other point of the circle is a right angle:
The exterior angle at a vertex equals the interior angle at the opposite vertex. Here is the whole argument in three lines.
Justification 1 — the chord formula.
Justification 2 — the triangle inequality. Let AB be a chord that does not pass through the centre O. Then A, O, B form a genuine triangle, and
Given: a point A inside a circle with centre O (A ≠ O). To show: among all chords through A, the shortest is the one perpendicular to OA.
Step 1 — chord length depends only on the distance from O. A chord at distance d from the centre has length 2√(r² − d²), which decreases as d increases. So the shortest chord is the one whose distance from O is greatest.
Step 2 — how large can that distance be? Let PQ be any chord through A and let M be the foot of the perpendicular from O to PQ, so the distance is OM.
Fig. 5.30 shows a semicircle on a diameter, with centre O, a point A on the arc, and the angles at the two ends of the diameter marked a and b. The tick marks tell you the key fact: the two halves of the diameter and the segment OA are all equal — they are all radii.
Call the ends of the diameter P and Q, so PO = OQ = OA = r. The segment OA splits the big triangle PAQ into two isosceles triangles.
Given: a circle with diameter AB; chords CC′ ⊥ AB and DD′ ⊥ AB. M is the midpoint of CD and M′ the midpoint of C′D′. To show: MM′ ⊥ AB.
Step 1 — AB is a mirror line for the whole figure. A diameter is a line of reflection symmetry of the circle. The chord CC′ is perpendicular to the diameter AB, so by Theorem 5 the diameter bisects it. Hence C′ is the reflection of C in the line AB. For the same reason D′ is the reflection of D in AB.
Step 2 — the midpoints are reflections of one another. Reflection is a rigid motion, so it sends the segment CD to the segment C′D′ and the midpoint of CD to the midpoint of C′D′. Therefore M′ is the reflection of M in AB. But the segment joining a point to its mirror image is always perpendicular to the mirror line. Hence MM′ ⊥ AB.
The same proof in coordinates. Take AB along the x-axis with the centre at the origin.
Fig. 5.31 shows a cyclic quadrilateral ABCD with the centre O joined to all four vertices. The tick marks say that OA, OB, OC, OD are all radii, so all four triangles OAB, OBC, OCD, ODA are isosceles. Four base angles are marked: p at A, q at B, u at C and v at D.
Now read off the four angles of the quadrilateral, each split by a radius into two of these pieces: