Diagonals AC and BD form right angles at point M in parallelogram ABCD. Prove ABCD is a rhombus.
2 answers:
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.
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.
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Answer:
41
Step-by-step explanation:
We know that complex numbers are a combination of real and imaginary numbers
Real part is x and imaginary part y is multiplied by i, square root of -1
Modulus of x+iy =
Here instead of x and y are given 9 and 40
i.e. 9+40i
Hence to find modulus we square the coefficients add them and then find square root
|9+49i| =
By long division method we find that
|9+40i| =41