prove that the diagonal of a rectangle bisect each other are equal.

1.	prove that the diagonal of a rectangle bisect each other are equal.
Answer» Let ABCD be a rectangle. Taking A as origin, the vertices of a rectangle are A(0, 0), B(a, 0), C(a, b) and D(0, b).\xa0Diagonal {tex}AC = \\sqrt {{{(a - 0)}^2} + {{(b - 0)}^2}} = \\sqrt {{a^2} + {b^2}} {/tex}Diagonal {tex}DB = \\sqrt {{{(a - 0)}^2} + {{(b - 0)}^2}} = \\sqrt {{a^2} + {b^2}} {/tex}{tex}\\therefore \\;AC = DB\\;\\left[ {\\because \\;{\\text{Each}}\\,\\sqrt {{a^2} + {b^2}} } \\right]{/tex}Mid-point of {tex}AC = \\left( {\\frac{{0 + a}}{2},\\frac{{0 + b}}{2}} \\right) = \\left( {\\frac{a}{2},\\frac{b}{2}} \\right){/tex}Mid-point of {tex}BD = \\left( {\\frac{{a + 0}}{2},\\frac{{0 + b}}{2}} \\right) = \\left( {\\frac{a}{2},\\frac{b}{2}} \\right){/tex}{tex}\\Rightarrow{/tex}\xa0Mid-point of AC = Mid-point of BD{tex}\\Rightarrow{/tex}\xa0Mid-point of AC and BD is same{tex}\\Rightarrow{/tex}\xa0AC and BD bisect each other.Hence the diagonals of a rectangles bisect each other and are equal proved.

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