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.
APPEARS IN
संबंधित प्रश्न
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
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.
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
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 such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
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. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
