Hong Kong
Stage 4 - Stage 5
Cyclic Quadrilaterals
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:
Prove that $\angle ABC$∠ABC = $\angle CDE$∠CDE.
By proving two similar triangles, Prove that $\angle BAD$∠BAD and $\angle DCE$∠DCE are equal.
Using this prove that $EB\times EC=ED\times EA$EB×EC=ED×EA.