Advertisements
Advertisements
Question
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
Advertisements
Solution
Given:
ABCD is a trapezium whose non-parallel sides AD and BC are equal.
Trapezium ABCD is cyclic.
Join BE, where BE || AD
Proof: Since, AB || DE and AD || BE
Since the quadrilateral ABED is a parallelogram.
∴ ∠BAD = ∠BED ...(i) [Opposite angles of a parallelogram are equal]
And AD = BE ...(ii) [Opposite angles of a parallelogram are equal]
But AD = BC [Given] ...(iii)
From equations (ii) and (iii)
BC = BE
⇒ ∠BEC = ∠BCE ...(iv) [Angles opposite to equal sides are equal]
Also, ∠BEC + ∠BED = 180° ...[Linear pair axiom]
∴ ∠BCE + ∠BAD = 180° ...[From equations (i) and (iv)]
If the sum of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.
Hence, trapezium ABCD is cyclic.
Hence proved.
RELATED QUESTIONS
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
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.
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.
In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.
Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.
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 in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
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 a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.
In the following figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.
