Advertisements
Advertisements
Question
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Advertisements
Solution
It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.
∠AEC + ∠CBA = 180°
∠AEC + ∠AED = 180° (Linear pair)
∠AED = ∠CBA ... (1)
For a parallelogram, opposite angles are equal.
∠ADE = ∠CBA ... (2)
From (1) and (2),
∠AED = ∠ADE
AD = AE (Angles opposite to equal sides of a triangle)
RELATED QUESTIONS
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
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?
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 trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
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.
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
