Question

In a figure, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E. Sides AD and BC are produced to meet at F. Angle ADC = 80 degree and angle BEC = 50 degree, then find angle BAD and angle CFD.

Answer 1

Answer:

∠BAD = 50°, ∠CFD = 30°

Step-by-step explanation:

In ΔADE, ∠ADE = ∠ADC = 80°                                     [given]

               ∠AED = ∠BEC = 50°                                      [given]

Now, ∠ADE + ∠AED + ∠EAD = 180°        

                                   [angle sum property of triangles]

        80° + 50° + ∠EAD = 180°

        130° + ∠EAD = 180°

       ∠EAD = 180° - 130°

       ∠EAD = 50°

Also, ABCD is a cyclic quadrilateral.

∠ADC + ∠ABC = 180°

                        [opp. angles of a cyclic quadrilateral are supplementary]

80° + ∠ABC = 180°

∠ABC = 180° - 80°

∠ABC = 100°

Similarly, ∠BCD = 130°

Now, ∠CDF = 180° - ∠CDA                   [linear pair]

         ∠CDF = 180° - 80°

         ∠CDF = 100°

Now, in ΔCDF, ∠BCD is an exterior angle.

⇒ ∠BCD = ∠CDF + ∠CFD

   130° = 100° + ∠CFD

  ∠CFD = 130° - 100°

  ∠CFD = 30°

                                                                            Hi! Hope this helped!!

Answer 2

Answer:

110

Step-by-step explanation: