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
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
In the given figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
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).
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
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 non-parallel sides of a trapezium are equal, prove that it is cyclic.
