As DE || BC, so, ∠ADE = ∠ABC

Also, ∠ABC = ∠ACB

∴ ∠ADE = ∠ACB                 (∵ AB = AC)

∴ ∠ADE + ∠EDB = 180°

⇒ ∠ACB + ∠EDB = 180°

Hence, B, C, D and E are concyclic.

OB = 4 cm

O’A = 2 cm

OO’ = 3 cm

As,       OO’ ≠ OB + O’A

So, circle does not touch each other externally.

Also,    OO’ ≠ OB – O’A

So, circle does not touch internally, hence they cut each other at two distinct points.

Here let O, O’ be the centers of the circle.

As the centre of each lies on the circumference of the other, the two circles will have the same radius. Let it be r.

∴ OC = O’C = r / 2

∴ AC =  r

AB = √ 3 r

Hence,  = √ 3 r : r = √ 3 : 1

∠OAC = ∠OCA

            2∠OAC = 80°            (External angle of ∆OAC)

            ∠OAC =  = 40°

⇒         x = 40°

Since ABCD is a cyclid trapezium then a cyclic trapezium being isosceles, so

∠DAB = ∠ABC and ∠ADC = ∠BCD

Let ∠DAB = x°

Then ∠BCD = 3x°

∴ x + 3x = 180° ⇒ x = 45°

∴ Largest angle = 3x = 3 × 45° = 135°

Let AB = 9 cm, BC = 7 cm, AC = 6 cm

Let x, y, z be radius of circles with centre A, B, C

 x + y = 9, y + z = 7 and z + x = 6

∴         2(x + y + z) = 22

Or        (x + y + z) = 11

∴         z = 11 – 9 = 2 cm

∠B + ∠D = 180°

⇒ ∠D = 180 – ∠B = 180° – 92° = 88°

∠DAE = ∠D = 88°

∠FAD = 88° + 20° = 108°

∠BCD = ∠FAD = 108°

⇒ ∠BCD = 108°

I. It is true that the opposite angles of a cyclic quadrilateral are supplementary.

II. It is also true that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Hence, both statements are individually true, but neither statements implies to each other.

Through four given points, we can draw at the most one circle.

OA = Diameter of inner circle = 4 cm

And OB = AB – OA = 6 – 4 = 2 cm

PQ and AB are two chord of outer circle.

∴         OA × OB = OP × OQ

                        (∵ OP = OQ)

⇒        4 × 2 = OP² ⇒ OP = 2 √ 2 cm

∴         PQ = 2 × 2 √ 2 = 4 √ 2 cm