Introduction, circle geometry

Circles i

1. Find the value of $x$ :

   

   

Theorem 4 The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circumference of the circle.

Proof :

Consider a circle, with centre $O$ and with $A$ and $B$ on the circumference. Draw a chord $AB$ . Draw radii $OA$ and $OB$ . Select any point $P$ on the circumference of the circle. Draw lines $PA$ and $PB$ . Draw $PO$ and extend to $R$ . The aim is to prove that $\widehat{AOB}=2·\widehat{APB}$ . $\widehat{AOR}=\widehat{PAO}+\widehat{APO}$ (exterior angle = sum of interior opp. angles) But, $\widehat{PAO}=\widehat{APO}$ ( $▵AOP$ is an isosceles $▵$ ) $\therefore$ $\widehat{AOR}=2\widehat{APO}$ Similarly, $\widehat{BOR}=2\widehat{BPO}$ . So,

Circles ii

1. Find the angles ( $a$ to $f$ ) indicated in each diagram:

   

Theorem 5 The angles subtended by a chord at the circumference of a circle on the same side of the chord are equal.

Proof :

Consider a circle, with centre $O$ . Draw a chord $AB$ . Select any points $P$ and $Q$ on the circumference of the circle, such that both $P$ and $Q$ are on the same side of the chord. Draw lines $PA$ , $PB$ , $QA$ and $QB$ . The aim is to prove that $\widehat{AQB}=\widehat{APB}$ .

Theorem 6 (Converse of Theorem [link] ) If a line segment subtends equal angles at two other points on the same side of the line, then these four points lie on a circle.

Proof :

Consider a line segment $AB$ , that subtends equal angles at points $P$ and $Q$ on the same side of $AB$ . The aim is to prove that points $A$ , $B$ , $P$ and $Q$ lie on the circumference of a circle. By contradiction. Assume that point $P$ does not lie on a circle drawn through points $A$ , $B$ and $Q$ . Let the circle cut $AP$ (or $AP$ extended) at point $R$ .

$\therefore$ the assumption that the circle does not pass through $P$ , must be false, and $A$ , $B$ , $P$ and $Q$ lie on the circumference of a circle.

Circles iii

1. Find the values of the unknown letters.

   

Cyclic quadrilaterals

Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle. The vertices of a cyclic quadrilateral are said to be concyclic .

Theorem 7 The opposite angles of a cyclic quadrilateral are supplementary.

Proof :

Consider a circle, with centre $O$ . Draw a cyclic quadrilateral $ABPQ$ . Draw $AO$ and $PO$ . The aim is to prove that $\widehat{ABP}+\widehat{AQP}={180}^{\circ }$ and $\widehat{QAB}+\widehat{QPB}={180}^{\circ }$ .

Theorem 8 (Converse of Theorem [link] ) If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

Proof :

Consider a quadrilateral $ABPQ$ , such that $\widehat{ABP}+\widehat{AQP}={180}^{\circ }$ and $\widehat{QAB}+\widehat{QPB}={180}^{\circ }$ . The aim is to prove that points $A$ , $B$ , $P$ and $Q$ lie on the circumference of a circle. By contradiction. Assume that point $P$ does not lie on a circle drawn through points $A$ , $B$ and $Q$ . Let the circle cut $AP$ (or $AP$ extended) at point $R$ . Draw $BR$ .

$\therefore$ the assumption that the circle does not pass through $P$ , must be false, and $A$ , $B$ , $P$ and $Q$ lie on the circumference of a circle and $ABPQ$ is a cyclic quadrilateral.