Question

1. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

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

Answer 1

Step-by-step explanation:

Given : ABCD is a cyclic quadrilateral, whose diagonals AC and BD are diameter of the circle passing through A, B, C and D. To Prove : ABCD is a rectangle. Proof : In ∆AOD and ∆COB AO = CO [Radii of a circle] OD = OB [Radii of a circle] ∠AOD = ∠COB [Vertically opposite angles] ∴ ∆AOD ≅ ∆COB [SAS axiom] ∴ ∠OAD = ∠OCB [CPCT] But these are alternate interior angles made by the transversal AC, intersecting AD and BC. ∴ AD || BC Similarly, AB || CD. Hence, quadrilateral ABCD is a parallelogram. Also, ∠ABC = ∠ADC ..(i) [Opposite angles of a ||gm are equal] And, ∠ABC + ∠ADC = 180° ...(ii) [Sum of opposite angles of a cyclic quadrilateral is 180°] ⇒ ∠ABC = ∠ADC = 90° [From (i) and (ii)] ∴ ABCD is a rectangle. [A ||gm one of whose angles is 90° is a rectangle] Proved.Read more on Sarthaks.com - https://www.sarthaks.com/12063/diagonals-cyclic-quadrilateral-diameters-circle-through-vertices-quadrilateral-rectangle