If ABCD is a quadrilateral inscribed in a unit circle whose side AB=1 and diagonal BD=√3what is the value of AD

Solution: 

AB = 1, BD = √3 , OA = OB = OD = 1

The given circle of radius 1 is also circumcircle of triangle ABD

∴ R = 1 for triangle ABD

∴ a/sinA = 2R

∴  √3/sinA = 2R = 2

∴ sinA = √3/2

∴ A = 60°

and hence C = 120o

Also by cosine rule on triangle ABD,

(√3)² = 1² + x² −2x cos60°

or   x² −x −2 = 0

∴ x = 2

Now, area ABCD = triangle ABD + triangle BCD

cd = 1 , ⇒ c²d² = 1

Also by cosine rule on triangle BCD we have

(√3)² = c² + d² − 2cd cos120°

= c² + d² + cd

∴ c² + d² = 2 or cd = 1

∴ c² and  d² are the roots of t² − 2t +1= 0

∴ c² =  d² = 1

∴ BC = 1 = CD and AD = x = 2

Solution: 

AB = 1, BD = √3 , OA = OB = OD = 1

The given circle of radius 1 is also circumcircle of triangle ABD

∴ R = 1 for triangle ABD

∴ a/sinA = 2R

∴  √3/sinA = 2R = 2

∴ sinA = √3/2

∴ A = 60°

and hence C = 120o

Also by cosine rule on triangle ABD,

(√3)² = 1² + x² −2x cos60°

or   x² −x −2 = 0

∴ x = 2

Now area ABCD = triangle ABD + triangle BCD

=3/4 –  ½ ( 1.2. sin60 ^{) + ( ½ .c.d. sin120}°)

= 3/4 – /2 + /4 c.d

cd = 1 , ⇒ c²d² = 1

Also by cosine rule on triangle BCD we have

(√3)² = c² + d² − 2cd cos120°

= c² + d² + cd

∴ c² + d² = 2 or cd = 1

∴ c² and  d² are the roots of t² − 2t +1= 0

∴ c² =  d² = 1

∴ BC = 1 = CD and AD = x = 2