Question

In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC//AE.If
∠ BAC=50°, find giving reasons:

(i) ∠ACB

(ii) ∠EDC

(iii) ∠BEC

Hence prove that BE is also a diameter

Answer 1

Let ∠ACB be x, ∠EDC be y and ∠BEC as z.
Now, ∠ABC = 90°                   {Angle in a semi-circle}
(i)  ∴ In ΔABC, ∠ABC + ∠BAC + ∠ACB = 180°

⇒ 90° + 50° + ∠ACB = 180°

        ∠ACB = 40° = x     

(ii) Now, ∠EAC = ∠ACB   {Alternate interior angles  ∴ AC is transversal to   the parallel lines AE & BC}

∴  ∠EAC = 40°

Also, ∠EAC + ∠EDC = 180°

∴ 40° + ∠EDC = 180°
⇒  ∠EDC  = 140° = y 

(iii)  ∠EBC + ∠EDC = 180°      {Angles in cyclic quadrilateral}

∠EBC + 140°  =  180° 
∠EBC = 40° 
∴ In ΔEBC
∠BEC + ∠ECB + ∠EBC = 180° 
∴ ∠BEC  + 90°  + 40°  = 180° 

∴ ∠BEC  = 50°

Also in  ΔEAB

∠EAB = ∠EAC + ∠BAC

          = 40° + 50°

∠EAB = 90°

We know, if an angle of a triangle in a circle is 90o then, the hypotenuse must be the diameter of the circle. Hence, BE is the diameter of the circle.