Question

please answer this question

Answer 1

ABCD is a rhombus
AD = CD
AE + ED = DF + FC But ED = FC
So , AE = DF
AB = AD
BE = AF
By SSS , ∆ABE is congruent to ∆DAF
So , a(BEA) = a(AFD)
180°-a(BED) = a(AFD)
So , a(AFD) + a(BED) = 180°