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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The value of the mean is 3 for normal curve C and the value of the mean is 4 for the normal curve D, and for both data standard deviation is the same.
<h3>What is a normal distribution?</h3>It's the probability curve of a continuous distribution that's most likely symmetric around the mean. On the Z curve, at Z=0, the chance is 50-50. A bell-shaped curve is another name for it.
We have a normal distribution curve shown in the picture.
For the normal curve C:
The value of mean = 3 (draw a vertical line in the middle of the curve)
For the normal curve D:
The value of mean = 4 (draw a vertical line in the middle of the curve)
For both data standard deviation is the same.
Thus, the value of the mean is 3 for normal curve C and the value of the mean is 4 for the normal curve D, and for both data standard deviation is the same.
Learn more about the normal distribution here:
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