Circle theorem proofs and linked geometry

3 exam-style questions. Answers and the mark for each step are on the last page.

Question 1Non-calculator · 4 marks
AB is a diameter of a circle with centre O. C is any point on the circle other than A or B. Join OC.
(a) Prove that angle ACB is a right angle without quoting the angle-in-a-semicircle theorem. (3)
(b) Why may the two base-angle pairs be taken equal? (1)
Question 2Non-calculator · 3 marks
$ABCD$ is a cyclic quadrilateral with the centre $O$ of the circle inside it. Angle $BAD = x$ and angle $BCD = y$ .
(a) Prove that $x + y = 180^\circ$ . You may use the fact that the angle at the centre is twice the angle at the circumference. (3)
Question 3Non-calculator · 3 marks
(a) A circle has centre O and radius 10 cm. P is outside the circle with OP = 26 cm. Tangents from P touch the circle at T and U. Work out the exact length of chord TU. You must show your working. (3)

Answers and marks

Question 1

(a) Let OAC=OCA=x and OBC=OCB=y, using equal radii. Then triangle ABC has angles x, y and x+y. Its angle sum gives 2x+2y=180, so ACB=x+y=90 degrees.

M1 Use OA=OC and OB=OC to establish both pairs of equal base angles.
M1 Write the full triangle angle sum using x and y.
C1 Correct conclusion with supporting reasoning: Let OAC=OCA=x and OBC=OCB=y, using equal radii. Then triangle ABC has angles x, y and x+y. Its angle sum gives 2x+2y=180, so ACB=x+y=90 degrees.

(b) Each relevant triangle has two radii as equal sides, so it is isosceles.

C1 Correct conclusion with supporting reasoning: Each relevant triangle has two radii as equal sides, so it is isosceles.

Question 2

(a) The two angles at $O$ on arcs $BD$ are $2x$ and $2y$ , which make $360^\circ$ .

M1 One angle at the centre written as $2x$ (or $2y$ ), with the correct arc.
M1 Both angles at the centre and the fact they add to $360^\circ$ .
C1 Dividing by 2 to conclude $x + y = 180^\circ$ with reasons given.

Question 3

(a) $\frac{240}{13}$ cm

P1 Establishing $\sqrt{26^{2}-10^{2}}$ or an equivalent valid method.
P1 Establishing $2\times 10\times 24/26$ or an equivalent valid method.
A1 Correct answer: $\frac{240}{13}$ cm

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Circle theorem proofs and linked geometry

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