Hello from MrBillDoesMath!
Answer:
No. That problem cited is one of 3 great unsolved problems of antiquity. See https://en.wikipedia.org/wiki/Angle_trisection for details.
Regards,
MrB
P.S. I'll be on vacation from Friday, Dec 22 to Jan 2, 2019. Have a Great New Year!
We compute for the side lengths using the distance formula √[(x₂-x₁)²+(y₂-y₁)²].
AB = √[(-7--5)²+(4-7)²] = √13
A'B' = √[(-9--7)²+(0-3)²] = √13
BC = √[(-5--3)²+(7-4)²] = √13
B'C' = √[(-7--5)²+(3-0)²] =√13
CD = √[(-3--5)²+(4-1)²] = √13
C'D' = √[(-5--7)²+(0--3)²] = √13
DA = √[(-5--7)²+(1-4)²] = √13
D'A' = √[(-7--9)²+(-3-0)²] = √13
The two polygons are squares with the same side lengths.
But this is not enough information to support the argument that the two figures are congruent. In order for the two to be congruent, they must satisfy all conditions:
1. They have the same number of sides.
2. All the corresponding sides have equal length.
3. All the corresponding interior angles have the same measurements.
The third condition was not proven.
Its a and b because it simply doesn't make any sense <span />
