Seg PQ is a chord of a circle. Arc PXQ is a major arc and Arc PYQ is a minor arc. ∠POQ is central angle of the circle :[Figure: a circle with centre O and a chord PQ; the major arc PXQ passes through the point X on the larger part of the circle and t…

Answer: (b) m∠PXQ = \(\dfrac{1}{2}\) m(arc PYQ), m∠PYQ = \(\dfrac{1}{2}\) m(arc PXQ), and m∠PXQ + m∠PYQ = 180º; (c) m∠POQ = m(arc PYQ) and m∠PXQ = \(\dfrac{1}{2}\) m∠POQ.

Note on the data: the question as given here carries no numerical measure. If the original figure shows a measure (for example of ∠POQ or of arc PYQ), substitute it in the relations below, as in the illustration at the end.

(a) Figure: draw a circle with centre O and a chord PQ; take X on the major arc and Y on the minor arc; join PX, XQ, PY, YQ, OP and OQ, and mark ∠PXQ, ∠PYQ and the central angle ∠POQ.

(b) ∠PXQ is an inscribed angle intercepting the minor arc PYQ, so m∠PXQ = \(\dfrac{1}{2}\) m(arc PYQ).

∠PYQ is an inscribed angle intercepting the major arc PXQ, so m∠PYQ = \(\dfrac{1}{2}\) m(arc PXQ).

Adding: m∠PXQ + m∠PYQ = \(\dfrac{1}{2}\)[m(arc PYQ) + m(arc PXQ)] = \(\dfrac{1}{2} \times 360º\) = 180º (the two arcs together form the whole circle; □PXQY is a cyclic quadrilateral, so opposite angles are supplementary).

(c) ∠POQ is the central angle of the minor arc PYQ, so m∠POQ = m(arc PYQ).

Hence m∠PXQ = \(\dfrac{1}{2}\) m∠POQ, i.e. m∠POQ = 2 m∠PXQ: the inscribed angle is half the central angle intercepting the same arc.

Illustration: if m∠POQ = 100º, then m(arc PYQ) = 100º, m(arc PXQ) = 360º − 100º = 260º, m∠PXQ = 50º, m∠PYQ = 130º, and 50º + 130º = 180º.