Question

c
11. In the adjoining figure, ABCD is a parallelogram, E is the mid-point of
AB and CE bisects BCD. Prove that:
(a) AE = AD
(1) DE bisects ZADC (a) DEC = 90
(Hint: BEC = ZECD = ZECB EB = BC = AE = AD.
ZADE = ZAED = EDC
A
E
B
LADC + ABCD = 180° =>
(SADC) + (LBCD) = 90° =ZEDC + ZDCE = 90°)​

Answer 1

Answer:

hello this is jannat thanks

Answer 2

Answer:

I hope this will help

Step-by-step explanation:

explanation is in the attachment