Question

prove that an isosceles trapezium is always cyclic and its diagonals are equal

Answer 1

Let ABCD be a cyclic quadrilateral and its diagonal AC and BD are the diameters of the circle through the vertices of the quadrilateral.



∠ABC = ∠BCD = ∠CDA = ∠DAB = 90° (Angles in the semi-circle)
Thus, ABCD is a rectangle as each internal angle is 90°.

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

Answer

Given,
ABCD is a trapezium where non-parallel sides AD and BC are equal.
Construction,
DM and CN are perpendicular drawn on AB from D and C respectively.
To prove,
ABCD is cyclic trapezium.



Proof:
In ΔDAM and ΔCBN,
AD = BC (Given)
∠AMD = ∠BNC (Right angles)
DM = CN (Distance between the parallel lines)
ΔDAM ≅ ΔCBN by RHS congruence condition.
Now,
∠A = ∠B by CPCT
also, ∠B

+

∠C = 180° (sum of the co-interior angles)
⇒ ∠A + ∠C = 180°
Thus, ABCD is a cyclic quadrilateral as sum of the pair of opposite angles is 180°.

Answer 2

Answer:

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