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 = [tex](\frac{x+x'}{2},\frac{y+y'}{2})[/tex]
coordinates of A and C are (0,0) and (2a + 2b, 2c)
Now mid point [tex]E=(\frac{2a+2b+0}{2},\frac{2c+0}{2})[/tex]
= [(a+b),c]
Now mid point of BD =[tex](\frac{2b+2a}{2},\frac{2c+0}{2})[/tex]
= [(b+a), c]
It proves that midpoints of AC and BD are common.
So AC trisects BD and BD bisects AC.
