(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 < QRS.
(b) ABCD is a cyclic quadrilateral and the diagonals AC and BD intersect at H. If < DAC = 41° and < AHB = 70°, calculate < ABC.
Explanation
(a)(i)
Given a circle APB with centre O.
To prove that < AOB = 2 < APB.
Construction: Draw PO and produce it to Q.
Proof : Referring to the lettering in the figure, |OA| = |OB| (radius)
(therefore x_{1} = x_{2}) (base angles of isosceles triangle)
(< AOQ = x_{1} + x_{2}) (exterior angles of triangle AOP)
(therefore < AOQ = 2x_{2} (x_{1} = x_{2}))
Similarly, (< BOQ = 2y_{2})
(< AOB = < AOQ + < BOQ )
Hence, (< AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2}))
= (2 times < APB)
(ii)
From the figure, (|OQ| = |QT|) (radius)
(< OQT = 22° = < QTO) (base angles of isosceles triangle)
(therefore < POQ = 44° ) (exterior angle of (Delta) QOT)
(< QOR = 180° – (44° + 98°) = 180° – 142° = 38°) (Angles on a straight line)
(< QOR = 2 times < QSR = 38°)
(therefore < QSR = frac{38°}{2} = 19°)
(b)
(< BHC = (180° – 70°) = 110°) (angle on a straight line)
(< DAC = 41° = < DBC) (angles on the same segment)
But (< DBC = < HBC)
(< AHB = < HBC + < HCB ) (exterior angles of (Delta) BHC)
(70° = 41° + < HCB implies < HCB = 70° – 41° = 29°)
(< ACB = < HCB = 29°)