Proof of the Circle Theorems: Practise Online

Either paper 5 marks

\(S\), \(T\) and \(U\) are points on a circle, centre \(O\).

\(ST\) is a diameter.

(a)
Explain why \(OS = OU\).

(b)
Explain why angle \(OUS = x\) and angle \(OUT = y\).

(c)
Hence prove that angle \(SUT = 90^\circ\).
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(a) \(OS\) and \(OU\) are both radii of the circle, so they are equal.
(b) Triangles \(OSU\) and \(OTU\) are isosceles, and base angles of an isosceles triangle are equal.
(c) The angles of triangle \(SUT\) give \(x + y + (x + y) = 180^\circ\), so \(x + y = 90^\circ\), and angle \(SUT = x + y\).
\(O\) is the centre, and \(S\) and \(U\) both lie on the circle, so \(OS\) and \(OU\) are both radii and therefore equal.
\(OS = OU\), so triangle \(OSU\) is isosceles and its two base angles are equal. The angle at \(S\) is \(x\), so the angle at \(U\) in that triangle is also \(x\).

\(OT = OU\) for the same reason, so triangle \(OTU\) is isosceles and the angle \(OUT\) equals the angle at \(T\), which is \(y\).
Angle \(SUT\) is made of the two pieces found in part (b), so it is \(x + y\).

The three angles of triangle \(SUT\) add up to \(180^\circ\):

\(x + y + (x + y) = 180^\circ\)

\(2(x + y) = 180^\circ\)

\(x + y = 90^\circ\)

So angle \(SUT = 90^\circ\), whatever the position of \(U\).
- (b) C1triangle \(OSU\) isosceles (\(OS = OU\)), so angle \(OUS = x\)
- (b) C1triangle \(OTU\) isosceles (\(OT = OU\)), so angle \(OUT = y\)
- (c) M1angles of triangle \(SUT\) sum to \(180^\circ\): \(x + y + (x + y) = 180^\circ\)
- (c) C1\(x + y = 90^\circ\), so angle \(SUT = 90^\circ\)
Either paper 5 marks

\(TA\) and \(TB\) are tangents to a circle with centre \(O\), touching the circle at \(A\) and at \(B\).

(a)
State the size of angle \(OAT\), and give a reason for your answer.

(b)
Prove that triangle \(OAT\) and triangle \(OBT\) are congruent.

(c)
Hence explain why \(TA = TB\).
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(a) \(90^\circ\). \(TA\) is a tangent and \(OA\) is the radius at the point of contact, so the angle between them is a right angle.
(b) RHS: both have a right angle, the hypotenuse \(OT\) is common, and \(OA = OB\) as radii.
(c) \(TA\) and \(TB\) are corresponding sides of congruent triangles, so they are equal.
\(TA\) is a tangent and \(OA\) is the radius drawn to the point where it touches, so the angle between them is \(90^\circ\).
Compare the two triangles.

Angle \(OAT\) and angle \(OBT\) are both \(90^\circ\), by part (a).

\(OT\) is a side of both triangles, and in each it is the hypotenuse.

\(OA = OB\), because both are radii of the circle.

A right angle, an equal hypotenuse and an equal side is the RHS condition, so the two triangles are congruent.
Congruent triangles have all their corresponding sides equal. \(TA\) and \(TB\) are corresponding sides of triangles \(OAT\) and \(OBT\), so \(TA = TB\).

That is the theorem: the two tangents drawn from a point outside a circle are equal in length.
- (b) B1a right angle in each triangle, where a tangent meets a radius
- (b) B1\(OT\) common and \(OA = OB\) as radii
- (b) B1congruent by RHS
Either paper 6 marks

\(E\), \(F\) and \(G\) are points on a circle, centre \(O\).

The line \(GO\) is continued to meet the circle again at \(H\).

(a)
Explain why angle \(OEG = x\).

(b)
Find the size of angle \(EOH\), in terms of \(x\).

(c)
Hence prove that angle \(EOF = 2 \times\) angle \(EGF\).
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(a) \(OE\) and \(OG\) are radii, so triangle \(OEG\) is isosceles and its base angles are equal.
(b) Angle \(EOH = 2x\)
(c) Angle \(FOH = 2y\) in the same way, so angle \(EOF = 2x + 2y = 2(x + y)\), and angle \(EGF = x + y\).
\(OE\) and \(OG\) are both radii of the circle, so they are equal and triangle \(OEG\) is isosceles.

The base angles of an isosceles triangle are equal, so the angle at \(E\) equals the angle at \(G\), which is \(x\).
Angle \(EOH\) is the exterior angle of triangle \(OEG\) at \(O\), so it equals the two opposite interior angles added together.

Those are the angle at \(G\) and the angle at \(E\), both \(x\).

\(x + x = 2x\)
Triangle \(OFG\) is isosceles for the same reason, so angle \(FOH = 2y\) by the argument in part (b).

Angle \(EOF\) is those two put together:

\(2x + 2y = 2(x + y)\)

Angle \(EGF\) is \(x + y\), because \(GO\) splits it into the two pieces marked \(x\) and \(y\).

So angle \(EOF = 2 \times\) angle \(EGF\).
- (a) M1\(OE\) and \(OG\) are both radii, so equal (triangle \(OEG\) isosceles)
- (a) C1base angles of an isosceles triangle are equal, so angle \(OEG = x\)
- (b) M1exterior angle of triangle \(OEG\) equals the sum of the two opposite interior angles, \(x + x\)
- (b) A1angle \(EOH = 2x\)
- (c) M1angle \(FOH = 2y\) in the same way, so angle \(EOF = 2x + 2y\)
- (c) A1\(2(x + y)\) and angle \(EGF = x + y\), completing the proof
Either paper 4 marks

\(W\), \(X\), \(Y\) and \(Z\) are points on a circle, centre \(O\).

\(Y\) and \(Z\) are on the same arc.

(a)
Write down \(z\) in terms of \(x\), and \(z\) in terms of \(y\).

Give a reason for your answers.

(b)
Hence prove that angle \(WYX = \) angle \(WZX\).
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(a) \(z = 2x\) and \(z = 2y\). Both \(Y\) and the centre stand on arc \(WX\), and the angle at the centre is twice the angle at the circumference.
(b) \(2x = z = 2y\), so \(x = y\), which means angle \(WYX = \) angle \(WZX\).
The angle at the centre is twice the angle at the circumference when both stand on the same arc.

\(Y\) and the centre both stand on arc \(WX\), so \(z = 2x\).

\(Z\) stands on the same arc, so \(z = 2y\) as well.
Both expressions in part (a) are equal to the same angle \(z\), so they are equal to each other.

\(2x = 2y\)

Dividing both sides by 2 gives \(x = y\).

So the two angles at \(Y\) and \(Z\) are equal — which is the theorem that angles in the same segment are equal.
- (a) B1\(z = 2x\) and \(z = 2y\)
- (a) C1reason: the angle at the centre is twice the angle at the circumference on the same arc
- (b) M1\(2x = z = 2y\), both equal to \(z\)
- (b) A1divide by 2 to get \(x = y\), so the angles at \(Y\) and \(Z\) are equal
Either paper 5 marks

\(W\), \(X\), \(Y\) and \(Z\) are points on a circle, centre \(O\).

(a)
Write down the angle \(XOZ\) on the same side of \(XZ\) as \(Y\), in terms of \(x\).

Give a reason for your answer.

(b)
Write down the other angle \(XOZ\), in terms of \(y\).

(c)
Hence prove that \(x + y = 180^\circ\).
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(a) \(2x\), because the angle at the centre is twice the angle at the circumference standing on the same arc \(XYZ\).
(b) The other angle \(XOZ\) is \(2y\), standing on arc \(XWZ\).
(c) The two angles at \(O\) make a full turn, so \(2x + 2y = 360^\circ\) and \(x + y = 180^\circ\).
The angle at \(W\) stands on arc \(XYZ\), and so does that angle at the centre.

The angle at the centre is twice the angle at the circumference, so it is \(2x\).
The angle at \(Y\) stands on the other arc \(XWZ\), and the angle at the centre standing on that arc is twice it, so \(2y\).
The two angles at \(O\) together make a complete turn about the centre.

\(2x + 2y = 360^\circ\)

\(2(x + y) = 360^\circ\)

\(x + y = 180^\circ\)

So opposite angles of a cyclic quadrilateral add up to \(180^\circ\).
- (a) B1\(2x\)
- (a) C1reason: angle at the centre is twice the angle at the circumference on the same arc
- (c) M1angles at \(O\) make a full turn, \(2x + 2y = 360^\circ\)
- (c) A1divide by 2 to get \(x + y = 180^\circ\)
Either paper 4 marks

\(P\), \(Q\) and \(R\) are points on a circle.

\(SP\) is a tangent to the circle.

\(QP = RP\)

Prove that \(y = 180^\circ - 2x\)

You must give a reason for each stage of your working.
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A reasoned chain ending at \(y = 180^\circ - 2x\): (1) angle \(PQR = x\), because the angle between a tangent and a chord (\(SPR\)) is equal to the angle in the alternate segment (\(PQR\)). (2) angle \(PRQ\) = angle \(PQR = x\), because \(PQ = PR\), so triangle \(PQR\) is isosceles and its base angles are equal. (3) \(y = 180^\circ - x - x\), because the angles in a triangle add up to \(180^\circ\). (4) so \(y = 180^\circ - 2x\).
Angle \(PQR = x\), because the angle between a tangent and a chord (\(SPR\)) is equal to the angle in the alternate segment (\(PQR\)).

Angle \(PRQ\) = angle \(PQR = x\), because \(PQ = PR\), so triangle \(PQR\) is isosceles and its base angles are equal.

\(y = 180^\circ - x - x\), because the angles in a triangle add up to \(180^\circ\).

So \(y = 180^\circ - 2x\).
- B1angle \(PQR = x\), because the angle between a tangent and a chord (\(SPR\)) is equal to the angle in the alternate segment (\(PQR\))
- B1angle \(PRQ\) = angle \(PQR = x\), because \(PQ = PR\), so triangle \(PQR\) is isosceles and its base angles are equal
- B1\(y = 180^\circ - x - x\), because the angles in a triangle add up to \(180^\circ\)
- A1so \(y = 180^\circ - 2x\)
Either paper 4 marks

\(P\), \(Q\) and \(R\) are points on a circle.

The straight line through \(P\) is a tangent to the circle.

The angle between the tangent and the chord \(PQ\) is \(y\).

Prove that angle \(PRQ = y\)

You may draw the diameter \(PS\).

You must not quote the alternate segment theorem. Give a reason for each stage of your working.
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Draw the diameter \(PS\). Angle between \(PS\) and the tangent \(= 90^\circ\) (tangent and radius), so angle \(QPS = 90^\circ - y\). Angle \(PQS = 90^\circ\) (angle in a semicircle). Angle \(PSQ = 180^\circ - 90^\circ - (90^\circ - y) = y\) (angles in a triangle add to \(180^\circ\)). Angle \(PRQ = \) angle \(PSQ = y\) (angles in the same segment are equal).
Draw the diameter \(PS\) and join \(Q\) to \(S\).

A tangent is perpendicular to the radius at the point of contact, so the diameter \(PS\) is at \(90^\circ\) to the tangent. The chord \(PQ\) lies between them and makes \(y\) with the tangent, so

\(\text{angle } QPS = 90^\circ - y\) (tangent and radius)

\(PS\) is a diameter, so

\(\text{angle } PQS = 90^\circ\) (angle in a semicircle)

The angles of triangle \(PQS\) add up to \(180^\circ\), so

\(\text{angle } PSQ = 180^\circ - 90^\circ - (90^\circ - y) = y\)

\(R\) and \(S\) are both on the same side of the chord \(PQ\), so they are in the same segment. Angles in the same segment are equal, so

\(\text{angle } PRQ = \text{angle } PSQ = y\)

which is what was to be proved.
- B1angle \(QPS = 90^\circ - y\), because the diameter \(PS\) passes through the centre, so it meets the tangent at \(90^\circ\) (tangent and radius)
- B1angle \(PQS = 90^\circ\), because it is the angle in a semicircle on the diameter \(PS\)
- M1angle \(PSQ = 180^\circ - 90^\circ - (90^\circ - y) = y\), from the angles of triangle \(PQS\) adding to \(180^\circ\)
- A1angle \(PRQ = y\), because \(R\) and \(S\) are in the same segment so angles \(PRQ\) and \(PSQ\) are equal; quoting the alternate segment theorem instead of this earns nothing
Either paper 4 marks

\(A\), \(B\) and \(C\) are points on the circumference of a circle, centre \(O\).

\(DB\) and \(DC\) are tangents to the circle.

\(AB = AC\)

Angle \(BDC = 2x\)

The diagram is not drawn accurately.

Prove that angle \(ABO = 45^\circ - \frac{x}{2}\)

You must give a reason for each stage of your working.
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angle \(BOC = 180^\circ - 2x\), angle \(ABC = 45^\circ + \frac{x}{2}\), angle \(OBC = x\), so angle \(ABO = 45^\circ - \frac{x}{2}\)
\(OB\) and \(OC\) are radii and \(DB\), \(DC\) are tangents, so angle \(OBD = OCD = 90^\circ\) (a tangent is perpendicular to the radius at the point of contact).

The angles in quadrilateral \(OBDC\) add up to \(360^\circ\), so

angle \(BOC = 360^\circ - 90^\circ - 90^\circ - 2x = 180^\circ - 2x\)

Triangle \(OBC\) is isosceles (\(OB = OC\), radii), so its base angles are equal:

angle \(OBC = (180^\circ - (180^\circ - 2x)) \div 2 = x\)

\(A\) is on the major arc \(BC\), so the arc \(BC\) that does not contain \(A\) subtends an angle of \(180^\circ - 2x\) at the centre. The angle at the centre is twice the angle at the circumference, so

angle \(BAC = 90^\circ - x\)

Triangle \(ABC\) is isosceles (\(AB = AC\)), so its base angles are equal:

angle \(ABC = (180^\circ - (90^\circ - x)) \div 2 = 45^\circ + \frac{x}{2}\)

\(O\) lies inside angle \(ABC\), so angle \(ABO\) is found by subtracting angle \(OBC\) from angle \(ABC\):

angle \(ABO = (45^\circ + \frac{x}{2}) - x = 45^\circ - \frac{x}{2}\)

So angle \(ABO = 45^\circ - \frac{x}{2}\).
- M1angles \(DBO\) and \(DCO\) are \(90^\circ\) (tangent perpendicular to radius), so angle \(BOC = 360^\circ - 90^\circ - 90^\circ - 2x = 180^\circ - 2x\) (angles in a quadrilateral)
- M1angle \(BAC = 90^\circ - x\) (angle at the centre is twice the angle at the circumference)
- M1angle \(ABC = 45^\circ + \frac{x}{2}\) (triangle \(ABC\) isosceles, \(AB = AC\)) and angle \(OBC = x\) (triangle \(OBC\) isosceles, radii)
- A1angle \(ABO\) found by subtracting the two angles and simplified to \(45^\circ - \frac{x}{2}\)
Either paper 5 marks

\(D\), \(E\) and \(F\) are points on a circle, centre \(O\).

\(DE\) is a diameter.

(a)
Explain why \(OD = OF\).

(b)
Explain why angle \(OFD = x\) and angle \(OFE = y\).

(c)
Hence prove that angle \(DFE = 90^\circ\).
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(a) \(OD\) and \(OF\) are both radii of the circle, so they are equal.
(b) Triangles \(ODF\) and \(OEF\) are isosceles, and base angles of an isosceles triangle are equal.
(c) The angles of triangle \(DFE\) give \(x + y + (x + y) = 180^\circ\), so \(x + y = 90^\circ\), and angle \(DFE = x + y\).
\(O\) is the centre, and \(D\) and \(F\) both lie on the circle, so \(OD\) and \(OF\) are both radii and therefore equal.
\(OD = OF\), so triangle \(ODF\) is isosceles and its two base angles are equal. The angle at \(D\) is \(x\), so the angle at \(F\) in that triangle is also \(x\).

\(OE = OF\) for the same reason, so triangle \(OEF\) is isosceles and the angle \(OFE\) equals the angle at \(E\), which is \(y\).
Angle \(DFE\) is made of the two pieces found in part (b), so it is \(x + y\).

The three angles of triangle \(DFE\) add up to \(180^\circ\):

\(x + y + (x + y) = 180^\circ\)

\(2(x + y) = 180^\circ\)

\(x + y = 90^\circ\)

So angle \(DFE = 90^\circ\), whatever the position of \(F\).
- (b) C1triangle \(ODF\) isosceles (\(OD = OF\)), so angle \(OFD = x\)
- (b) C1triangle \(OEF\) isosceles (\(OE = OF\)), so angle \(OFE = y\)
- (c) M1angles of triangle \(DFE\) sum to \(180^\circ\): \(x + y + (x + y) = 180^\circ\)
- (c) C1\(x + y = 90^\circ\), so angle \(DFE = 90^\circ\)
Either paper 5 marks

\(TS\) and \(TT\) are tangents to a circle with centre \(O\), touching the circle at \(S\) and at \(T\).

(a)
State the size of angle \(OST\), and give a reason for your answer.

(b)
Prove that triangle \(OST\) and triangle \(OTT\) are congruent.

(c)
Hence explain why \(TS = TT\).
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(a) \(90^\circ\). \(TS\) is a tangent and \(OS\) is the radius at the point of contact, so the angle between them is a right angle.
(b) RHS: both have a right angle, the hypotenuse \(OT\) is common, and \(OS = OT\) as radii.
(c) \(TS\) and \(TT\) are corresponding sides of congruent triangles, so they are equal.
\(TS\) is a tangent and \(OS\) is the radius drawn to the point where it touches, so the angle between them is \(90^\circ\).
Compare the two triangles.

Angle \(OST\) and angle \(OTT\) are both \(90^\circ\), by part (a).

\(OT\) is a side of both triangles, and in each it is the hypotenuse.

\(OS = OT\), because both are radii of the circle.

A right angle, an equal hypotenuse and an equal side is the RHS condition, so the two triangles are congruent.
Congruent triangles have all their corresponding sides equal. \(TS\) and \(TT\) are corresponding sides of triangles \(OST\) and \(OTT\), so \(TS = TT\).

That is the theorem: the two tangents drawn from a point outside a circle are equal in length.
- (b) B1a right angle in each triangle, where a tangent meets a radius
- (b) B1\(OT\) common and \(OS = OT\) as radii
- (b) B1congruent by RHS