Diagonals ac and bd form right angles at point m in parallelogram abcd. prove abcd is a rhombus.

1. Consider the parallelogram ABCD drawn in the 1. figure attached.

2. The diagonals of a parallelogram bisect each other, so AM=MC and BM=MD

3. In triangle ABC, BM is an altitude (BM perpendicular to AC), but also a median (AM=MC). An altitude is also a median only when the triangle is isosceles, so AB=BC 

similarly, in triangle BCD, CM is both an altitude and a median.Thus, BCD is an isosceles triangle so BC=CD.

4. AB= DC (opposite sides of a parallelogram are equal), similarly BC=DA

5. So ABCD is a quadrilateral, with all 4 sides equal in length. Thus ABCD is a rhombus. 

1.) 18+5 is one way i’m pretty sure there are others but i can’t think of them right now

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