Proof of the Circle Theorems
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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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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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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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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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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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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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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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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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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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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) \(OS\) 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 \(SUT = x + y\). -
(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. -
(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 \(EGF = x + y\). -
(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 \(WZX\). -
(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 \(x + y = 180^\circ\). - 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 \(y = 180^\circ - 2x\).
- 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).
- angle \(BOC = 180^\circ - 2x\), angle \(ABC = 45^\circ + \frac{x}{2}\), angle \(OBC = x\), so angle \(ABO = 45^\circ - \frac{x}{2}\)
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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 \(DFE = x + y\). -
(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.
Takes a couple of seconds the first time a sheet is made.