OPPOSITE ANGLES OF A CYCLIC QUADRILATERAL ARE SUPPLEMENTARY

A quadrilateral is called cyclic quadrilateral if all its four vertices lie on the circumference of the circle. Now we are going to learn the special property of cyclic quadrilateral.

Theorem :

Opposite angles of a cyclic quadrilateral are supplementary.

Proof : 

Consider the quadrilateral ABCD whose vertices lie on a circle. We want to show that its opposite angles are supplementary.

Connect the center O of the circle with each vertex. You now see four radii OA, OB, OC and OD giving rise to four isosceles triangles OAB, OBC, OCD and ODA.

The sum of the angles around the center of the circle is 360°. The angle sum of each isosceles triangle is 180°. 

2(∠1 + ∠2 + ∠3 + ∠4) + Angle at center O = 4  180°

2(∠1 + ∠2 + ∠3 + ∠4) + 360° = 720°

Subtract 360° from each side. 

2(∠1 + ∠2 + ∠3 + ∠4) = 360°

Divide each side by 2.

∠1 + ∠2 + ∠3 + ∠4 = 180°

You now interpret this as


(i) (∠1 + ∠2) + (∠3 + ∠4) = 180° 

(sum of opposite angles B and D)

(i) (∠1 + ∠4) + (∠2 + ∠3) = 180° 

(sum of opposite angles A and C)

Converse of the Theorem :

If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.

Example 1 :

In the diagram shown below. ABCD is a cyclic quadrilateral in which BCD = 100° and ABD = 50°. Find ADB.

Solution :

By Theorem,

BAD + ∠BCD = 180°

Substitute BCD = 100°.

BAD + 100° = 180°

Subtract 100° from each side. 

∠BAD = 80°

In triangle ABD,

∠ABD + ∠BAD + ADB = 180°

50° + 80° + ADB = 180°

130° + ADB = 180°

Subtract 130° from each side.

ADB = 50°

Example 2 :

In the diagram shown below, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ∠CBD = 30° and BAC = 50°Find CAD and BCD.

Solution :

∠CAD :

The angles at the circumference subtended by the same arc are equal. ∠CAD and ∠CBD at the circumference subtended by the same arc CD.

∠CAD = ∠CBD

∠CAD = 30°

∠BCD :

By Theorem,

∠BAD + ∠BCD = 180°

In the above diagram, ∠BAD = ∠BAC + ∠CAD.

(∠BAC + ∠CAD) ∠BCD = 180°

∠BAC + ∠CAD ∠BCD = 180°

Substitute ∠BAC = 50° and ∠CAD = 30°.

50° + 30° ∠BCD = 180°

80° ∠BCD = 180°

Subtract 80° from each side. 

∠BCD = 100°

Example 3 :

In the diagram shown below, O is the center of the circle and ADC = 120°. Find the value of x.

Solution :

By Theorem,

ABC + ADC = 180°

Substitute ∠ADC = 120°.

ABC + 120° = 180°

Subtract 120° from each side. 

ABC = 60°

∠ACD = 90° (angle in a semi circle is a right angle)

In triangle ABC,

BAC + ACB + ABC = 180°

x° + 90° + 60° = 180°

x + 90 + 60 = 180

x + 150 = 180

Subtract 150 from each side.

x = 30

Example 4 :

In the figure given below, ABCD is a cyclic quadrilateral in which AB || DC. If BAD = 100° find BCD, ADC and ABC.

Solution :

∠BCD :

By Theorem,

BAC + ∠BCD = 180°

Substitute BAD = 100°.

100° + BCD = 180°

Subtract 100° from each side.

∠BCD = 80°

∠ADC :

Since AB || DC, AD is transversal.

When two parallel lines intersected by a transversal, same side interior angles are supplementary. 

∠BAD + ∠ADC = 180°

Substitute BAD = 100°.

100° + ∠ADC = 180°

Subtract 100° from each side.

∠ADC = 80°

∠ABC :

By Theorem, 

ABC + ∠ADC = 180°

Substitute ADC = 80°.

∠ABC + 80° = 180°

Subtract 80° from each side. 

∠ABC = 100°

Example 5 :

In the diagram shown below, PQ is a diameter of a circle with center O. If PQR = 55°SPR = 25° and PQM = 50°. Find QPR, QPM and PRS.

Solution :

QPR : 

PRQ  =  90° (angle in a semi circle is a right angle)

In triangle PRQ,

PRQ + QPR + PQR = 180°

90° + QPR + 55° = 180°

QPR + 145° = 180°

Subtract 145° from each side.

QPR = 35°

QPM : 

In triangle QPM,

QPM + MQP + QMP = 180°

QPM + 50° + 90° = 180°

QPM + 140° = 180°

Subtract 140° from each side. 

QPM = 40°

∠PRS : 

By Theorem,

PQR + PSR = 180°

55° + PSR = 180°

Subtract 55° from each side. 

PSR = 125°

In triangle PSR,

PSR + SPR + PRS = 180°

125° + 25° PRS = 180°

150° + PRS = 180°

Subtract 150° from each side. 

PRS = 30°

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