Question

The midpoint of AC is

Answer 1

Answer:

Step-by-step explanation:

Given ABCD is a parallelogram

To prove ⇒ AC trisects BD, and BD bisects AC.

Proof ⇒ Since coordinates of points A, B, C and D have been given in the diagram. If we prove that midpoint of AC and BD are common then AC and BD will equally bisect each other.

Midpoint of AC = $(\frac{x+x'}{2},\frac{y+y'}{2})$

coordinates of A and C are (0,0) and (2a + 2b, 2c)

Now mid point  $E=(\frac{2a+2b+0}{2},\frac{2c+0}{2})$

                             = [(a+b),c]

Now mid point of BD =$(\frac{2b+2a}{2},\frac{2c+0}{2})$

                                   = [(b+a), c]

It proves that midpoints of AC and BD are common.

So AC trisects BD and BD bisects AC.

Answer 2

The midpoint of $AC$ is $E$.

Hope this helps.

r3t40