Theory question 108
Circle Theorems(a)(i) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference. (ii) In the diagram above, O is the centre of the circle and PT is a diameter. If ∠PTQ = 22° and ∠TOR = 98°, calculate ∠QSR. (b) ABCD is a cyclic quadrilateral and the diagonals AC and BD intersect at H. If ∠DAC = 41° and ∠AHB = 70°, calculate ∠ACB.
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(a)(i) Arc AB, centre O, P on the remaining circumference. Join PO and produce to N.
OA = OP (radii) ⇒ ∠OAP = ∠OPA = x; exterior angle ∠AON = 2x.
Similarly ∠BON = 2y where y = ∠OPB.
∠AOB = 2x + 2y = 2(x + y) = 2∠APB.
(ii) ∠PTQ = 22° ⇒ arc PQ subtends 44° at O.
∠POT = 180° (diameter), so ∠QOR = 180° − 44° − 98° = 38°.
∠QSR = ½ × 38° = 19°.
(b) ∠DBC = ∠DAC = 41° (angles in the same segment).
∠BHC = 180° − 70° = 110° (angles on a straight line).
In △BHC: ∠ACB = 180° − 110° − 41° = 29°.
Use 'angles in the same segment are equal' to move the 41° to B.