The answer
(a) similar by AA
(b)(i) proven
(b)(ii) proven
O-Level A-Math 2025 Paper 2 Question 10 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 10 of the O-Level A-Math 2025 Paper 2. It tests alternate segment theorem, in the Proofs in plane geometry area. It is worth 10 marks: (a) 3 + (b)(i) 3 + (b)(ii) 4. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
(a) In triangles \(ABE\) and \(BCE\): - \(\angle AEB = \angle BEC\) (the same angle at \(E\), since \(A\), \(C\), \(E\) are collinear); - \(\angle BAE = \angle CBE\) (\(\angle BAE = \angle BAC\) is the angle in the alternate segment to the tangent-chord angle \(\angle CBE\), by the alternate segment theorem).
With two pairs of equal angles, triangle \(ABE\) is similar to triangle \(BCE\) (AA), with correspondence \(A \leftrightarrow B\), \(B \leftrightarrow C\), \(E \leftrightarrow E\).
(b)(i) The tangent \(FBE\) is a straight line, so \(\angle FBA + \angle ABC + \angle CBE = 180^{\circ}\). By the alternate segment theorem \(\angle CBE = \angle BAC = x\), and \(\angle FBA = y\), so \(\angle ABC = 180^{\circ} - x - y\). Since \(ABCD\) is a cyclic quadrilateral, opposite angles are supplementary: \(\angle ADC = 180^{\circ} - \angle ABC = 180^{\circ} - (180^{\circ} - x - y) = x + y\). (proven)
(b)(ii) If \(DC \parallel AB\), then with transversal \(AC\), \(\angle BAC = \angle DCA\) (alternate angles). \(\angle BAC\) is the inscribed angle standing on arc \(BC\), and \(\angle DCA\) is the inscribed angle standing on arc \(AD\). Equal inscribed angles stand on equal arcs, so arc \(BC =\) arc \(AD\), and equal arcs subtend equal chords. Therefore \(AD = BC\). (proven)
Answer: (a) similar by AA
(b)(i) proven
(b)(ii) proven
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Our O-Level A-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a alternate segment theorem question from Proofs in plane geometry, worth 10 marks: (a) 3 + (b)(i) 3 + (b)(ii) 4.
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