Lesson

A cyclic quadrilateral is a four-sided shape that has all its vertices touching the circle's circumference, such as the one shown below.

The Theorem

The opposite angles in a cyclic quadrilateral add up to $180^\circ$180°.

Proof:

 $ABCD$ABCD is a cyclic quadrilateral (given) Join $AC$AC to $BD$BD
 $\angle CAB+\angle ABC+\angle ACB$∠CAB+∠ABC+∠ACB $=$= $180^\circ$180° (angle sum of a triangle) $\angle CAB$∠CAB $=$= $\angle CDB$∠CDB (angles in the same segment of a circle are equal) $\angle ACB$∠ACB $=$= $\angle ADB$∠ADB (angles in the same segment of a circle are equal)

Therefore, adding the previous two statements we get

 $\angle ACB+\angle CAB=\angle ADB+\angle CDB$∠ACB+∠CAB=∠ADB+∠CDB $=$= $\angle ADC$∠ADC $\angle ACB+\angle CAB+\angle ABC$∠ACB+∠CAB+∠ABC $=$= $180^\circ$180° then - Adding $\angle ABC$∠ABC on both sides $\angle ACB+\angle CAB+\angle ABC$∠ACB+∠CAB+∠ABC $=$= $180^\circ$180° (Angle sum of a triangle) $\angle ADC+\angle ABC$∠ADC+∠ABC $=$= $180^\circ$180° $\angle BAD+\angle BCD=360^\circ-\left(\angle ADC+\angle ABC\right)$∠BAD+∠BCD=360°−(∠ADC+∠ABC) $=$= $180^\circ$180°

Remember!

The converse of this theorem is also true.

If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.

#### Worked Examples

##### Question 1

In the diagram, $O$O is the centre of the circle. Show that $x$x and $y$y are supplementary angles.

##### Question 2

Consider the figure:

1. Prove that $\angle ABC$ABC = $\angle CDE$CDE.

2. By proving two similar triangles, Prove that $\angle BAD$BAD and $\angle DCE$DCE are equal.

3. Using this prove that $EB\times EC=ED\times EA$EB×EC=ED×EA.