Mathematics
In the adjoining figure; AB = AD, BD = CD and ∠DBC = 2∠ABD. Prove that : ABCD is a cyclic quadrilateral.
Answer
In △ABD,
AB = AD
∠ADB = ∠ABD (∵ angles opposite to equal sides are equal) ……(1)
In △BDC,
BD = CD
∠DCB = ∠DBC (∵ angles opposite to equal sides are equal) ……(2)
In △ADB,
⇒ ∠DAB + ∠ADB + ∠ABD = 180° [Angle sum property of triangle]
⇒ ∠DAB + ∠ABD + ∠ABD = 180° [From (1)]
⇒ ∠DAB + 2∠ABD = 180°
⇒ ∠DAB + ∠DBC = 180° [As, ∠DBC = 2∠ABD (Given)]
⇒ ∠DAB + ∠DCB = 180° [From (2)]
Since, ∠DAB and ∠DCB are opposite angles of a quadrilateral and sum of opposite angles in a cyclic quadrilateral = 180°.
Hence, proved that ABCD is a cyclic quadrilateral.
Related Questions
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In the given figure, AB and XY are diameters of a circle with center O. If ∠APX = 30°, find :
(i) ∠AOX
(ii) ∠APY
(iii) ∠BPY
(iv) ∠OAX
AB is a diameter of a circle with centre O. Chord CD is equal to radius OC. AC and BD produced intersect at P. Prove that : ∠APB = 60°.
In the given figure, AC = AB and ∠ABC = 72°. OA and OB are two tangents. Determine :
(i) ∠AOB
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