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Referred to an origin O, the position vectors of four non-collinear points A, B, C and D are a, b, c and d respectively. Given that a - b = d - c, show that ABCD is a parallelogram. Given also that |a-c|=|b-d| identify the shape of the parallelogram ABCD, justifying your answer.​

User Gskema
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Answer:

Hence, the shape of the parallelogram ABCD is a rectangle.

Explanation:

We know that a - b = d - c.

Let E be the midpoint of AC. Then we have:

E = (a + c)/2

Similarly, let F be the midpoint of BD. Then we have:

F = (b + d)/2

Using these midpoints, we can express the position vectors of B and D as follows:

b = 2E - a

d = 2F - c

Substituting these expressions into the given equation, we get:

a - (2E - a) = (2F - c) - c

Simplifying this equation, we get:

2a - 2E = 2F - 2c

a - c = F - E

This means that the vector joining A and C is equal to the vector joining F and E. Thus, AC is parallel to FE.

Similarly, we can show that BD is parallel to FE. Hence, ABCD is a parallelogram.

Now, we are given that |a-c| = |b-d|. Since the opposite sides of a parallelogram are equal in length, we have:

AB = CD = |b - a|

BC = AD = |c - b|

Therefore, we have AB = CD and BC = AD. This means that the opposite sides of ABCD are equal in length. Also, we have shown that AC is parallel to BD. Therefore, ABCD is a parallelogram with opposite sides equal in length and parallel to each other.

Hence, the shape of the parallelogram ABCD is a rectangle.

User Wvd
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