CHECK IF THE GIVEN QUADRILATERAL IS CYCLIC OR NOT

Is ABCD a cyclic quadrilateral ? Explain your answers.

Example 1  :

Solution  :

By observing the figure, m<B and m<D are opposite angles.

So,

m<B + m<D  =  107˚ + 73˚

m<B + m<D  =  180˚ (supplementary angles)

So, there is one pair of opposite angles that are supplementary.

Yes, it is a cyclic quadrilateral.

Example 2  :

Solution  :

In the figure, m<B  =  m<c  =  47˚

So, m<A and m<D subtends equal angles at <B and <C

Yes, it is a cyclic quadrilateral.

Example 3  :

Solution  :

By observing the figure, m<B and m<D are opposite angles.

So,

m<B + m<D  =  87˚ + 87˚

m<B + m<D  =  174˚ (not supplementary angles)

There is one pair of opposite angles that are not supplementary.

So, it is not a cyclic quadrilateral.

Example 4  :

Solution  :

By observing the figure, it is a rectangle (all the angles are 90˚).

Then, m<B and m<D are opposite angles.

m<B + m<D  =  90˚ + 90˚

m<B + m<D  =  180˚(supplementary angles)

So, opposite angles are supplementary.

Yes, it is a cyclic quadrilateral.

Example 5  :

Solution  :

In the figure, m<A and m<E  =  113˚(interior and exterior angle)

The sum of the interior angle of a triangle is 180˚

 

180˚ - <C  =  113˚

180˚ - 113˚  =  <C

<C  =  67˚

Now, m<A and m<C are opposite angles.

m<A + m<C  =  113˚ + 67˚

m<A + m<C  =  180˚ (supplementary angles)

So, there is one pair of opposite angles that are supplementary.

Yes, it is a cyclic quadrilateral.

Example 6 :

Solution :

By vertically opposite angles,

<BFC  =  <AFD  =  80˚

<DFC  =  <AFB (adjacent angle)

We know that vertical angle and its adjacent angle are supplementary angles)

<BFC + <DFC  =  180˚

80˚ + <DFC  =  180˚

<DFC  =  180˚ - 80˚

=  100˚

<DFC  =  <AFB  =  100˚

By linear pair of angles,

<EDB  =  115˚

<EDB + <CDB  =  180˚

<CDB  =  180˚ - <EDB

<CDB  =  180˚ - 115˚

<CDB  =  65˚

In ∆DFC,

<FDC + <DFC + <DCF  =  180˚

100˚ + 65˚ + <DCF  =  180˚

<DCF  =  180˚ - 165˚

<DCF  =  15˚

We know that the angle subtended by the same arc are equal.

<CDB  =  <CAB  =  65˚

<DCA  =  <DBA  =  15˚

m<B  =  m<c  =  15˚

So, m<A and m<D subtends equal angles at <B and <C

Yes, it is a cyclic quadrilateral.

Example 7 :

ABCD is a cyclic parallelogram. Show that it is a rectangle.

Solution :

Given that ABCD is a cyclic quadrilateral, then sum of opposite angles will be equal to 180^o.

∠A + ∠C = 180^o ----(1)

Since it is parallelogram, the opposite sides are equal and opposite angles are also equal.

∠A = ∠C

applying in (1), we get

∠A + ∠A = 180^o

2∠A = 180^o

∠A = 180^o / 2

= 90^o

So, the given is a rectangle.

Example 8 :

A pair of opposite sides of a cyclic quadrilateral is equal. Prove that its diagonals are also equal

Solution :

Let ABCD be a cyclic quadrilateral and AB = CD.

arc AB = arc CD (Corresponding arcs)

Adding arc AD to both the sides;

arc AB + arc AD = arc CD + arc AD

∴ arc BAD = arc CDA

Chord BD = Chord CA ⇒ BD = CA

Example 9 :

In Figure PQRS is a cyclic quadrilateral whose diagonals intersect at A. If ∠SQR = 80^o and ∠QPR = 30^o, find ∠SRQ.

Solution :

∠SQR = 80^o and ∠QPR = 30^o

∠SQR = 80^o = ∠SPR

Angles created by the same arc PS

∠QPR = 30^o = ∠QSR

Angles created by the same arc QR

In triangle SQR,

∠QSR + ∠SQR + ∠SRQ = 180

30 + 80 + ∠SRQ = 180

110 + ∠SRQ = 180

∠SRQ = 180 - 110

∠SRQ = 70

Example 10 :

PQRS is a cyclic quadrilateral. If ∠Q = ∠R = 65^o , find ∠P and ∠S.

∠P + ∠R = 180^o

∠P = 180^o – ∠R

= 180^o – 65^o

∠P = 115^o

Similarly, ∠Q + ∠S = 180^o

∠S = 180^o – ∠Q

= 180^o – 65^o

∠S = 115^o

Example 11 :

In Figure, ABCD is a cyclic quadrilateral whose diagonals intersect at O. If ACB = 50^o and ∠ ABC = 110^o, find ∠ BDC.

Solution :

In triangle ABC,

∠BAC + ∠ABC + ∠BCA = 180

∠BAC + 110 + 50 = 180

∠BAC + 160 = 180

∠BAC = 180 - 160

∠BAC = 20

∠BDC = 20

Angles created by the same arc BC.

Example 12 :

In Figure, PQRS is a cyclic quadrilateral, and the side PS is extended to the point A. If ∠PQR = 80^o, find ∠ASR.

Solution :

Since the given shape is a cyclic quadrilateral,

∠PQR +  ∠PSR = 180

80 + ∠PSR = 180

∠PSR = 180 - 80

∠PSR = 100

∠ASR and ∠PSR are supplementary.

Then ∠ASR = 80

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.