Cyclic Quadrilateral
A quadrilateral having its all four
vertices on the circumference of a circle is called a cyclic quadrilateral. ABCD,
in the given figure is a cyclic quadrilateral.
Concyclic Points
The points lying on the circumference of
the circle are called concyclic points. In the given circle, A, B, C, D are
concyclic points.
Circle Theorems on Cyclic Quadrilateral
1.
The sum of opposite angles of
a cyclic quadrilateral is always 180°.
2.
In a cyclic quadrilateral, exterior
angle is equal to its opposite interior angle.
Proofs:
1. The sum of opposite
angles of a cyclic quadrilateral is always 180°.
Given: PQRS is a cyclic quadrilateral.
To Prove: (i) ∠QPS + ∠QRS = 180°
(ii) ∠PQR + ∠PSR = 180°
Proof:
Statements Reasons
1.
∠QPS = ½ Arc QRS -------> Inscribed angle is equal to the half
of its opposite arc.
2.
∠QRS = ½ Arc QPS -------> Inscribed angle is equal to the half
of its opposite arc.
3.
∠QPS + ∠QRS = ½ Arc QRS + ½ Arc QPS -----> Adding statements 1 and 2.
or, ∠QPS + ∠QRS = ½ Arc (QRS + QPS)
or, ∠QPS + ∠QRS = ½ Circle PQRS
or, ∠QPS + ∠QRS = ½ 360°
or, ∠QPS + ∠QRS = 180°
4.
∠PQR + ∠PSR = 180° -----------> Same as above.
Proved.
2. In a cyclic
quadrilateral, exterior angle is equal to its opposite interior angle.
Given: ABCD is a cyclic quadrilateral with exterior angle ∠DCE.
To Prove: ∠DCE = ∠BAD
Proof:
Statements Reasons
1.
∠DCE + ∠BCD = 180° -----> Opposite angles of a cyclic quadrilateral
are supplementary.
2.
∠BAD + ∠BCD = 180° -----> Linear pair of angles.
3.
∠DCE + ∠BCD = ∠BAD + ∠BCD ------> From statements 1 and 2.
4.
∠DCE = ∠BAD ---------> Cancelling common angle ∠BCD.
Proved.
Look at the following worked out examples
on circle and cyclic quadrilateral:
Workout Examples
Example 1: Find ∠COD in the given figure.
Solution: From the figure,
110° + ∠ODC = 180° [Opposite angles of a cyclic quadrilateral]
or, ∠ODC
= 180° - 110°
or, ∠ODC
= 70°
∠ODC
= ∠OCD = 70° [Being OC = OD, radii of same circle]
∠COD + ∠ODC + ∠OCD = 180° [Sum of angles of DCOD]
or, ∠COD + 70° + 70° = 180°
or, ∠COD + 140° = 180°
or, ∠COD = 180° - 140°
or, ∠COD = 40°
Example 2: Calculate the value of x and y in the given figure.
Solution: From the figure,
x = ½ ∠BOD [Inscribed angle is half of central angle]
or, x = ½ 160°
or, x = 80°
y + x = 180° [Opposite angles of a
cyclic quadrilateral]
or, y + 80° = 180°
or, y = 180° - 80°
or, y = 100°
Hence, x = 80° and y = 100°
Example 3: Find the angle x in the given figure.
Solution: From the figure,
∠CAD
= 56° [Inscribed angle on same arc CD]
∠BAD
= 96° [Exterior angle of a cyclic quadrilateral]
x = ∠BAD - ∠CAD
= 96° - 56°
= 40°
Example 4: In the adjoining figure, ABCD is a cyclic quadrilateral,
side CD is produced to the point E, where BC = DE. If CA bisects ∠BCD prove that DACE is an isosceles
triangle.
Solution:
Given:
ABCD is a cyclic quadrilateral. BC = DE and ∠BCA
= ∠ACD.
To Prove:
DACE is an isosceles triangle.
Proof:
Statements Reasons
1. ∠BCA = ∠ACD ---------------> Given.
2. Arc AB = arc AD ------------>
Arcs subtended by equal inscribed angles.
3. AB =AD ---------------> Equal
chords cut off equal arcs in a circle.
4. In DABC and DADE
i. AB
= AD (S) --------> From statement 3
ii. ∠ABC = ∠ADE (A) -------> Exterior angle of a cyclic quadrilateral
iii. BC
= DE (S) -------> Given.
5. DABC ≅ DADE -----------> By SAS axiom
6. AC = AE -------------> Corresponding sides of ≅ triangles.
7. ACE is an isosceles triangle ----------> Being AC = AE.
Proved.
Example 5: In the given figure, SP is the bisector of ∠RPT and PQRS is a cyclic
quadrilateral. Prove that SQ = SR.
Solution:
Given:
PQRS is a cyclic quadrilateral. ∠RPS
= ∠SPT.
To Prove:
SQ = SR.
Proof:
Statements Reasons
1.
∠SPT = ∠RPS -----------> From given.
2.
∠SPT = ∠QRS -----------> Exterior angle of cyclic quadrilateral
PQRS.
3.
∠SQR = ∠RPS -----------> Inscribed angle on same arc RS.
4.
∠SQR = ∠QRS
----------> From statements 1, 2 and 3.
5.
SQ = SR -------> Being base angles of DSQR equal, statement 4.
Proved.
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