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

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

the two may have that, even a breakdown is probably offered