Prove that angle ADC is congruent to angle CBA

with solution!

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

lo.lll

step-by-step explanation:

- 70

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