UK Secondary (7-11)
topic badge
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:

  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.

 

 

 

 

 

 

What is Mathspace

About Mathspace