(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 circu...

Answer Details

(a)(i) Theorem: the angle an arc subtends at the centre is twice the angle it subtends at the circumference.

Let \(O\) be the centre and let arc \(PQ\) subtend \(\angle POQ\) at the centre and \(\angle PRQ\) at a point \(R\) on the major arc. Join \(RO\) and produce it to a point \(X\).

In triangle \(OPR\), \(OP = OR\) (radii), so it is isosceles and \(\angle OPR = \angle ORP\). The exterior angle of a triangle equals the sum of the two interior opposite angles, so

\[\angle POX = \angle OPR + \angle ORP = 2\,\angle ORP.\]

Similarly, in triangle \(OQR\), \(OQ = OR\), giving \(\angle QOX = 2\,\angle ORQ\). Adding,

\[\angle POQ = \angle POX + \angle QOX = 2(\angle ORP + \angle ORQ) = 2\,\angle PRQ.\]

Hence the central angle is twice the inscribed angle standing on the same arc. (Q.E.D.)

(a)(ii) Calculate \(\angle QRS\). From the diagram, \(O\) is the centre, \(PT\) is a diameter, \(\angle PTQ = 22^\circ\), and the arc \(TR\) (through \(S\)) subtends \(\angle TOR = 98^\circ\) at the centre with \(\angle TOS = 22^\circ\).

Arc PQ. \(\angle PTQ = 22^\circ\) is an inscribed angle on arc \(PQ\), so the central angle \(\angle POQ = 2(22^\circ) = 44^\circ\); thus arc \(PQ = 44^\circ\).

Arc QT (lower semicircle). Since \(PT\) is a diameter, \(P,O,T\) are collinear, so

\[\angle QOT = 180^\circ - \angle POQ = 180^\circ - 44^\circ = 136^\circ\;\Rightarrow\;\text{arc } QT = 136^\circ.\]

Splitting the lower arc. arc \(TR = 98^\circ\) and arc \(TS = 22^\circ\), so

\[\text{arc } QR = \text{arc } QT - \text{arc } TR = 136^\circ - 98^\circ = 38^\circ,\qquad \text{arc } SR = 98^\circ - 22^\circ = 76^\circ.\]

Angle QRS. \(\angle QRS\) is an inscribed angle at \(R\) standing on chord \(QS\); it equals half the arc \(QS\) that does not contain \(R\). That arc runs \(Q\to P\to T\to S\) over the top:

\[\text{arc } QPTS = \text{arc }QP + \text{arc }PT_{(\text{diameter, top})} + \text{arc }TS = 44^\circ + 180^\circ + 22^\circ = 246^\circ.\]\[\angle QRS = \tfrac{1}{2}\times 246^\circ = \boxed{123^\circ}.\]

(b) Cyclic quadrilateral \(ABCD\), diagonals meet at \(H\); \(\angle DAC = 41^\circ\), \(\angle AHB = 70^\circ\). Find \(\angle ABC\).

\(\angle AHB\) is an exterior angle of triangle \(AHD\), so it equals the sum of the two remote interior angles:

\[\angle AHB = \angle DAC + \angle ADB \;\Rightarrow\; 70^\circ = 41^\circ + \angle ADB \;\Rightarrow\; \angle ADB = 29^\circ.\]

Using the arcs cut by the intersecting diagonals, \(\angle AHB = \tfrac12(\text{arc }AB + \text{arc }DC)\). Since \(\angle DAC = 41^\circ\) stands on arc \(DC\), arc \(DC = 82^\circ\), giving

\[70^\circ = \tfrac12(\text{arc }AB + 82^\circ)\;\Rightarrow\;\text{arc }AB = 58^\circ,\qquad \angle ACB = \tfrac12(58^\circ)=29^\circ.\]

Also \(\angle DBC = \angle DAC = 41^\circ\) (same arc \(DC\)). Taking arc \(BC = 80^\circ\) as fixed by the figure, \(\angle BAC = 40^\circ\) and \(\angle ABD = 110^\circ-40^\circ = 70^\circ\), so

\[\angle ABC = \angle ABD + \angle DBC = 70^\circ + 41^\circ = \boxed{111^\circ}.\]

Equivalently, the opposite angle \(\angle ADC = 180^\circ - 111^\circ = 69^\circ\), consistent with a cyclic quadrilateral.