Question

In the following figure. ABCD is a cyclic
quadrilateral in which AD is parallel to BC.
If the bisector of angle A meets BC at point E
and the given circle at point F. prove that:
(1) EF = FC (ii) BF - DF​

Answer 1

Given : ABCD is a cyclic quadrilateral with AD || BC and AF is the angle bisector of ∠A which intersect BC of E  

Let ∠BAF = ∠FAD = x

We know that angles in the same segment are equal

⇒ ∠BDF = ∠BAF = x  ... (1)

and ∠DBF = ∠FAD = x  ... (2)

  Now in ∆BDF we have

∠BDF = ∠DBF = x              (from (1) and (2))

⇒ BF = DF                (sides opposite to equal angles are equal)  ...  (3)

 Also we know that in a cyclic quadrilateral. If two sides are parallel then its other two sides are equal

⇒ AB = DC  ...  (4)

 

Now in ∆BEF and ∆DCF

∠FBE = ∠CDF                (Angles in the same segment)

BF = DF              (from (3))

and ∠BFE = ∠DFC             (Angles subtended by equal chords are equal)

Thus ∆BEF ≅ ∆DCF                  (by ASA congruency criterion)

⇒ EF = CF                   (C.P.C.T.).