Question

In the following  Figure  ABCD is a  arallelogram, CE bisects ∠C and AF bisects ∠A. In each of the following, if the statement is true, give a reason for the same:

(i) ∠A = ∠C
(ii) \[\angle FAB = \frac{1}{2}\angle A\] 

(iii) \[\angle DCE = \frac{1}{2}\angle C\]

(iv) \[\angle CEB = \angle FAB\]

(v) CE || AF 

Answer 1

(i) True, since opposite angles of a parallelogram are equal.
(ii) True, as AF is the bisector of\[\angle\] A. 

(iii) True, as CE is the bisector of \[\angle\]C. 

(iv) True

\[\angle\]CEB =\[\angle\] DCE........(i)  (alternate angles) 

\[\angle\]DCE= \[\angle\] FAB.........(ii) (opposite angles of a parallelogram are equal)
            
  From equations (i) and (ii):

\[\angle\] CEB =\[\angle\]FAB 

(v) True, as corresponding angles are equal (\[\angle\] CEB =\[\angle\] FAB).