Advertisements
Advertisements
प्रश्न
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Advertisements
उत्तर

Consider a trapezium ABCD with AB | |CD and BC = AD.
Draw AM ⊥ CD and BN ⊥ CD.
In ΔAMD and ΔBNC,
AD = BC (Given)
∠AMD = ∠BNC (By construction, each is 90°)
AM = BN (Perpendicular distance between two parallel lines is same)
∴ ΔAMD ≅ ΔBNC (RHS congruence rule)
∴ ∠ADC = ∠BCD (CPCT) ... (1)
∠BAD and ∠ADC are on the same side of transversal AD.
∠BAD + ∠ADC = 180° ... (2)
∠BAD + ∠BCD = 180° [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.
संबंधित प्रश्न
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

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.
Prove that a cyclic parallelogram is a rectangle.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
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.

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
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.
