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 opposite angles in a cyclic quadrilateral add up to $$180°.
$$ABCD | is a cyclic quadrilateral (given) | |
Join | $$AC to $$BD |
$$∠CAB+∠ABC+∠ACB | $$= | $$180° | (angle sum of a triangle) |
$$∠CAB | $$= | $$∠CDB | (angles in the same segment of a circle are equal) |
$$∠ACB | $$= | $$∠ADB | (angles in the same segment of a circle are equal) |
Therefore, adding the previous two statements we get
$$∠ACB+∠CAB=∠ADB+∠CDB | $$= | $$∠ADC | |
$$∠ACB+∠CAB+∠ABC | $$= | $$180° | then - Adding $$∠ABC on both sides |
$$∠ACB+∠CAB+∠ABC | $$= | $$180° | (Angle sum of a triangle) |
$$∠ADC+∠ABC | $$= | $$180° | |
$$∠BAD+∠BCD=360°−(∠ADC+∠ABC) | $$= | $$180° |
The converse of this theorem is also true.
If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.
In the diagram, $$O is the centre of the circle. Show that $$x and $$y are supplementary angles.
Consider the figure:
Prove that $$∠ABC = $$∠CDE.
By proving two similar triangles, Prove that $$∠BAD and $$∠DCE are equal.
Using this prove that $$EB×EC=ED×EA.