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OK, so for 34A I have to prove that the parallelogram MATH IS A RHOMBUS, IT SAYS TO FIND THAT ALL SIDES ARE CONGRUENT, BUT IN A SQUARE ALL SIDES ARE CONGRUENT and when I try to prove perpendicular bisection, that’s another property. So I’m lost. It would be much appreciated if someone could help me understand what the question means for the first part because I think I’m overthinking this.

OK so for 34A I have to prove that the parallelogram MATH IS A RHOMBUS IT SAYS TO FIND THAT ALL SIDES ARE CONGRUENT BUT IN A SQUARE ALL SIDES ARE CONGRUENT and class=


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