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
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°-A, 90° − `1/2 A, 90° − 1/2 B, 90° − 1/2` C.
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.
ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.
