Step-by-step explanation:
< i'll use this too represent angle
this is only presuming that this is a parallelogram and i also don't know what those numbers mean
< ACB = < CAD (alt <s of DA II CB) A
< ACD = < CAB (alt <s of DC II AB) A
AB is common S
therefore, ∆ADC is congruent to ∆CBA
if ∆ADC is congruent to ∆CBA
then < ADC = < CBA
