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Nezavi [6.7K]
3 years ago
12

Diagonals AC and BD form right angles at point M in parallelogram ABCD. Prove ABCD is a rhombus.

Mathematics
2 answers:
Tanya [424]3 years ago
5 0
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. 

BabaBlast [244]3 years ago
5 0

The correct answer is:

B. △AMB ≅ △CMB ≅ △CMD ≅ △AMD

|Huntrw6|

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This works!

So the integers are -7 and -6.
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