Cyclic Quadrilateral
A quadrilateral having its all four vertices on the circumference of a circle is called a cyclic quadrilateral.
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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.
Theorems on Cyclic Quadrilateral and Their Proofs:
THEOREM 1:
The sum of opposite angles of a cyclic quadrilateral is always
180°.
PROOFS:
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 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.
Hence proved.
THEOREM 2:
In a cyclic quadrilateral, the exterior angle is equal to its
opposite interior angle.
PROOFS:
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.
Hence proved.
Problems on Cyclic Quadrilaterals
Worked Out 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° Ans.
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°
Ans.
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]
Now,
x = ∠BAD – ∠CAD
= 96° – 56°
= 40°
Ans.
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: Here,
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.
Hence proved.
Example 5: In the given figure, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = SR.
Solution: Here,
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.
Hence proved.
If you have any question or problems regarding the Cyclic
Quadrilaterals, you can ask here, in the comment
section below.
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