Question

Prove that the diagonals of a rectangle are equal and bisect each other.

Answer 1

let ABCD be a rectangle, diagonal AC and BD intersect at the point o
from triangle ABC and BADwe have
AB=BA
angle ABC = angle BAD (both is equal to 90 degree.
BC=AD ( opposite sides of rectangle )
now triangle ABC congruence triangle BAD
AC =BD
hence the diagonal of rectangle are equal.
this shows the diagonal of rectangle are bisect each other

Answer 2

${\underline{\underline{\frak{Answer\::}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀☯ Let OABC be a rectangle such that OA is a along x - axis and OB is along y - axis.

⠀⠀⠀⠀⠀⠀⠀⠀☯ Let OA = a and OB = b.

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\setlength{\unitlength}{1.7cm}\begin{picture}(6, 6)\thicklines\put(0, 0){\vector(0, 1){3}}\put(0, 0){\vector( 0, - 1){2}}\put(0, 0){\vector(1, 0){3}}\put(0, 0){\vector( - 1, 0){2}}\put(0.1,-0.3){\sf O(0,0)}\put(1.4,-0.3){\sf A(a,0)}\multiput(1.7,0)(0,0.2){7}{\line(0,2){0.1}}\multiput(0, 1.4)(0.2,0){9}{\line(2,0){0.1}}\put(1.4,1.6){\sf C(a,b)}\put( - 0.7,1.4){\sf B(o,b)}\put( - 2.3,-0.1){ \sf X \:\'{} }\put(3.2,-0.1){\sf X}\put(-0.1,3.2){\sf Y}\put(-0.1,-2.3){ \sf Y\:\'{} }\end{picture}$

☯ Then, the coordinates of A and B are (a,0) and (0,b) respectively.

Since, OABC is a rectangle. Therefore,

$:\implies\sf AC = OB$

$:\implies\sf AC = b$

Thus, we have

$:\implies\sf AC = a\;and\;AC = b$

So, the coordinates of C are (a,b).

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

✇ The coordinates of the mid - point of OC are,

$:\implies\sf \bigg( \dfrac{a + 0}{2} , \dfrac{b + 0}{2} \bigg) = \bigg( \dfrac{a}{2} , \dfrac{b}{2} \bigg)$

✇ Also, The coordinates of the mid - point of AB are,

$:\implies\sf \bigg( \dfrac{a + 0}{2} , \dfrac{b + 0}{2} \bigg) = \bigg( \dfrac{a}{2} , \dfrac{b}{2} \bigg)$

Clearly, coordinates of the mid - point of OC and AB are same.

Hence, OC and AB bisect each other.

Also,

$:\implies\sf \pink{OC = \sqrt{a^2 + b^2}}$

and,

$:\implies\sf AB = \sqrt{(a - 0)^2 + (0 - b)^2}$

$\qquad:\implies\sf \purple{AB = \sqrt{a^2 + b^2}}$

Therefore, OC = AB

$\therefore$ Hence, Diagonal of a rectangle bisect each other and are equal.