
Patrick B. answered 07/14/20
Math and computer tutor/teacher
I am very sorry that I am having an extremely difficult time following this proof.
So here is q quick summary of the proof...
Parallelogram ABCD, with A in the top left going clockwise...
M = AC intersect BD
It shall suffice to show that angle BAD and able CDA are equal and supplementary.
By SSS:
triangle BAD = triangle DCB
triangle BAD = triangle CDA
By CPCTC:
angle BAD = angle DCB
angle BAD = angle CDA
which by transitive/substitution:
angle DCB = angle BAD = angle CDA
BUT.... angles CDA and BAD are SUPPLEMENTARY
(and at the same time equal, as just proven)
so therefore they must both be 90 degrees.