Prove that: “In a parallelogram, the diagonals bisect each other” by using vectors
To prove that in a parallelogram, the diagonals bisect each other using vectors, we can utilize the properties of parallelograms and vector arithmetic.
Let's consider a parallelogram with vertices A, B, C, and D. We can denote the position vectors of these points as vector A, vector B, vector C, and vector D, respectively.
Now, the midpoint of a line segment AB can be found by taking the average of the position vectors of points A and B.
So, the midpoint of diagonal AC (denoted as point M) can be found using the formula:
M = (A + C)/2
Similarly, the midpoint of diagonal BD (denoted as point N) can be found using:
N = (B + D)/2
To prove that the diagonals bisect each other, we need to show that the point M is the same as the point N.
To do this, let's compute the position vectors of points M and N using the given formulas:
M = (A + C)/2 = (1/2)A + (1/2)C
N = (B + D)/2 = (1/2)B + (1/2)D
To show that M and N are equal, we can set their position vectors equal to each other and solve for the unknowns:
(1/2)A + (1/2)C = (1/2)B + (1/2)D
To simplify this equation further, we can multiply both sides by 2:
A + C = B + D
Note that A and B represent the same vector since they correspond to the opposite sides of the parallelogram, and similarly, C and D represent the same vector.
Hence, we can write:
A + C = B + D as A - B = D - C
Since the left side is the negative of the right side, this shows that vector D - C is equal in magnitude and opposite in direction to vector A - B.
Thus, we have established that the diagonals bisect each other since the point M and N coincide, implying that the vector connecting the midpoints of the diagonals is zero.