Question

Quadrilateral ABCD is a cyclic quadrilateral Ray AP bisects Angle B A D Ray C Q bisects angle BCD.
To prove: PQ is a diameter of the circle.​

Answer 1

Answer:ji bilkul yaad ho muje aap aapse bhi baat ki hh mene but but uske baad aapni baat hui hi ni isliye

its yogita

Answer 2

Given:

ABCD is a cyclic quadrilateral

AP bisects ∠BAD

CQ bisects ∠BCD

To Prove:

PQ is the diameter of the circle.

Solution:

∠BDC = ∠BAC ( angles in the same segment )

⇒ ∠BDC = 30°

               [ ∵ ∠BAC = 30° ( given ) ]

⇒ In ΔBCD

         ∠BDC + ∠DBC + ∠BCD  = 180°

                               [ Angle sum property ]

⇒ 30° + 70° + ∠BCD = 180°

                      [ ∠BCD = 70° , ∠BDC = 30° ]

∠BCD = 180° - 30° - 70°

          = 80°

If AB = BC,

           Then ∠BCA = ∠BAC = 30°

           [ Angles opp. to equal sides in a Δ are equal ]

Now ∠ECD = ∠BCD - ∠BCE

                   = 80° - 30°

                   = 50°

Hence, ∠BCD = 80° and ∠ECD = 50°