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Stage 4 - Stage 5

Lesson

In Quadrilaterals in Circles, we looked at cyclic quadrilaterals. A cyclic quadrilateral is a four-sided shape that has all its vertices touching the circle's circumference. We learnt that:

• the opposite angles in a cyclic quadrilateral add up to $180^\circ$180°, and conversely,
• if a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.

However, sometimes we are asked to prove that these features are true.

Watch out!

If you're asked to prove a rule is true, you can't use that rule in your proof.

So how do we do this?

There are several different proofs of the cyclic quadrilateral property. Here is one of them.

### Proof

We construct the diagonals of a cyclic quadrilateral and label with the same letter the angles that are known to be equal due to the fact that angles at the circumference subtended by the same arc are equal. In any of the triangles formed by a diagonal, we have $a+b+c+d=180^\circ$a+b+c+d=180° since this is the angle sum of a triangle. We can associate the components of this sum in the following two ways to show that the opposite angles of the cyclic quadrilateral are supplementary.

$(a+b)+(c+d)=180^\circ$(a+b)+(c+d)=180°
$(a+c)+(b+d)=180^\circ$(a+c)+(b+d)=180°.

We must also show that if the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic. To prove this, we assume that it is not necessarily true that a quadrilateral whose opposite angles sum to $180^\circ$180° is cyclic and we obtain a contradiction that shows the assumption to be false. In the diagram, the quadrilateral $PQRS$PQRS has been constructed so that the opposite angles sum to $180^\circ$180°. The triangle $PQS$PQS has a unique circumcircle, that is, a circle that passes through each of its vertices. We assume that vertex $R$R of the quadrilateral does not lie on this circumcircle.

The edge $QR$QR of the quadrilateral cuts the circumcircle at $T$T, making a cyclic quadrilateral $QTSP$QTSP. By the first part of the proof, we must have $\angle P+\angle QTS=180^\circ$P+QTS=180°. But, by the construction, we have $\angle P+\angle R=180^\circ$P+R=180°

It follows that $\angle QTS=\angle R$QTS=R so that lines $ST$ST and $SR$SR must be parallel. This means points $R$R and $T$T must coincide and thus, $PQRS$PQRS is a cyclic quadrilateral after all.

Having confirmed the theorem about cyclic quadrilaterals, we can use it and the rest of our geometrical knowledge, including our knowledge of properties of quadrilaterals, congruent triangles, as well as circle geometry, to construct further proofs.

Some Circle Geometry Rules
• Chords of equal lengths subtend equal angles at the centre.
• Angles subtended by the same arc are equal.
• The angle inscribed in a semicircle is always a right angle.

#### Example

Find a pair of similar triangles in the following diagram and show that $RS\times RP=RT\times RQ$RS×RP=RT×RQ. The quadrilateral $PQTS$PQTS is cyclic. Therefore, $\angle Q+\angle TSP=180^\circ$Q+TSP=180°. But, $\angle TSP+\angle TSR=180^\circ$TSP+TSR=180°. So, $\angle TSR=\angle Q$TSR=Q.

In the triangles $RST$RST and $RQP$RQP, $\angle R$R is common to both. Therefore, triangles $RST$RST and $RQP$RQP are similar because they have the same angles.

We can form ratios of corresponding sides to obtain $\frac{RS}{RQ}=\frac{RT}{RP}$RSRQ=RTRP. It follows from this that $RS\times RP=RT\times RQ$RS×RP=RT×RQ.

#### Worked Examples

##### Question 1

Prove that $x=y$x=y. ##### 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.

##### Question 3

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