Question

Prove that the diagonals of a rectangle ABCD, with vertices A$(2,-1)$, B$(5,-1)$, C $(5,6)$ and D $(2,6)$, are equal and bisect each other.

Answer 1

$\Delta ADC$ and $\Delta BDC$ are right angled triangle with AD and BC are hypotenuse.

$A{C}^2=A{B}^2+D{C}^2$

$A{C}^2=(5-2{)}^2+(6+1{)}^2=9+48=58$ sq.unit

$B{D}^2=D{C}^2+C{B}^2$

$B{D}^2=(5-2{)}^2+(-1-6{)}^2=9+49=58$ sq.unit

Hence, both the diagonals are equal in length.

In $\Delta ABO$ and $\Delta CDO$

Since, $\angle OAB=\angle OCD$, $\angle OBA=\angle ODC$ (Both are alternate interior angles of parallel lines)

and $AB=CD$

Therefore $\Delta ABO\cong \Delta CDO$

$\Rightarrow AO=CO$ and $BO=DO$

Therefore, Both diaginals bisects each other.