In Fig. 2.15, points A, B, C, D and E are collinear such that AB = BC = CD = DE. Then
(a) AD = AB + ______
(b) AD = AC + ______
(c) midpoint of AE is ______
(d) midpoint of CE is ______
(e) AE = ______ × AB.

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Solution:-
(a) AD = AB + BD
Where, BD = BC + CD
(b) AD = AC + CD
(c) midpoint of AE is c
(d) midpoint of CE is D
(e) AE = 4 × AB. [given]







