In Fig. 17.31, ABCD is a parallelogram, CE bisects ∠C and AF bisects ∠A.

Question:

In Fig. 17.31, ABCD is a parallelogram, 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) $\angle F A B=\frac{1}{2} \angle A$

(iii) $\angle D C E=\frac{1}{2} \angle C$

(iv) $\angle C E B=\angle F A B$

(v) CE || AF

Solution:

(i) True, since opposite angles of a parallelogram are equal.

(ii) True, as AF is the bisector of ">A.

(iii) True, as CE is the bisector of ">C.

(iv) True

">">CEB = ">DCE........(i)  (alternate angles)

">">DCE= "> FAB.........(ii)    (opposite angles of a parallelogram are equal)

 From equations (i) and (ii):">

">CEB = ">FAB

(v) True, as corresponding angles are equal (">CEB = ">FAB).

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