Question

prove that the diagonals of rectangle the vertices (0,0),(a,0),(a,b),(0,b) bisect each other and are equal?

Answer 1

Let A (0,0) B (a,0) C (a, b) and D (0, b)
here AC and BD is diagonal of rectangle .
midpoint of AC
use section formula
{(0+a)/2, (0+b)/2}=(a/2, b/2)

also find midpoint of BC
{(a+0)/2,(0+b)/2}=(a/2, b/2)

we see
midpoint of AC =midpoint of BD
hence midpoint of both diagonals meet at a same point (a/2, b/2)
so, we can say that diagonals of a rectangle bisect each other are equal.


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Answer 2

let A (0,0) B (a,0) C (a, b) and D (0, b)

here AC and BD is diagonal of rectangle .

midpoint of AC

use section formula

{(0+a)/2, (0+b)/2}=(a/2, b/2)

also find midpoint of BC

{(a+0)/2,(0+b)/2}=(a/2, b/2)

we see

midpoint of AC =midpoint of BD

hence midpoint of both diagonals meet at a same point (a/2, b/2)

so, we can say that diagonals of a rectangle bisect each other are equal.