Advertisements
Advertisements
Question
If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.
Advertisements
Solution
Given: Let ABCD be a cyclic quadrilateral and AD = BC.
Join AC and BD.
To prove: AC = BD
Proof: In ΔAOD and ΔBOC,
∠OAD = ∠OBC and ∠ODA = ∠OCB ...[Since, same segments subtends equal angle to the circle]
AB = BC ...[Given]
ΔAOD = ΔBOC ...[By ASA congruence rule]
Adding is DOC on both sides, we get
ΔAOD + ΔDOC ≅ ΔBOC + ΔDOC
⇒ ΔADC ≅ ΔBCD
AC = BD ...[By CPCT]
APPEARS IN
RELATED QUESTIONS
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
ABCD is a cyclic quadrilateral in ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =

