OPPOSITE ANGLES OF A CYCLIC QUADRILATERAL ARE SUPPLEMENTARY PROOF
Opposite angles of a cyclic quadrilateral are supplementary (or) The sum of opposite angles of a cyclic quadrilateral is 180°
Given : O is the centre of circle. ABCD is the cyclic quadrilateral.
To prove : ∠BAD + ∠BCD = 180°, ∠ABC + ∠ADC = 180°
Construction : Join OB and OD
(i) ∠BAD = (1/2)∠BOD.
(The angle substended by an arc at the centre is double the angle on the circle.)
(ii) ∠BCD = (1/2) reflex ∠BOD.
(iii) ∠BAD + ∠BCD = (1/2)∠BOD + (1/2) reflex ∠BOD.
Add (i) and (ii).
∠BAD + ∠BCD = (1/2)(∠BOD + reflex ∠BOD)
∠BAD + ∠BCD = (1/2) ⋅ (360°)
(Complete angle at the centre is 360°)
∠BAD + ∠BCD = 180°
(iv) Similarly ∠ABC + ∠ADC = 180°.
Problem 1 :
In the figure given below, O is the center of a circle and ∠ADC = 120°. Find the value of x.
ABCD is a cyclic quadrilateral. we have
∠ABC + ∠ADC = 180°
∠ABC = 180° - 120°
∠ABC = 60°
Also ∠ACB = 90° (angle on a semi circle).
In triangle ABC we have,
∠BAC + ∠ACB + ∠ABC = 180°
∠BAC + 90° + 60° = 180°
∠BAC = 180° - 150°
∠BAC = 30°
So, the value of x is 30.
Problem 2 :
In the figure given below, ABCD is a cyclic quadrilateral in which AB || DC. If ∠BAD = 100° find.
∠BAD + ∠BCD = 180°
100° + ∠BCD = 180°
∠BCD = 180° - 100°
∠BCD = 80°
Because AB || DC and AD is tranversal,
∠BAD + ∠ADC = 180°
100° + ∠ADC = 180°
Subtract 180° from each side.
∠ADC = 80°
∠ADC + ∠ABC = 180°
80° + ∠ABC = 180°
∠ABC = 180° - 80°
∠ABC = 100°
Problem 3 :
In the figure given below, ABCD is a cyclic quadrilateral in which ∠BCD = 100° and ∠ABD = 50° find ∠ADB.
∠DAB + ∠DCB = 180°
∠DAB + 100° = 180°
∠DAB = 180° - 100°
∠DAB = 80°
In triangle ADB,
∠DAB + ∠ABD + ∠ADB = 180°
80° + 50° + ∠ADB = 180°
130° + ∠ADB = 180°
Subtract 130° from each side.
∠ADB = 180°
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