Question 697261


Let the diagonals intersect at {{{O}}}

{{{AB = CD}}} sides of rectangles are congruent

< {{{ABO}}}= < {{{CDO}}} alternate angles {{{AB || BC}}}
< {{{BAO}}} = < {{{DCO}}} alternate angles {{{AB || BC}}}

so triangles {{{AOB}}} and {{{COD}}} are {{{congruent}}}.

then {{{AO = OC}}}

But {{{AC = AO + OC= 2AO}}}

{{{AO = (1/2) AC}}}

similarly you can prove that {{{BO = DO}}}

the diagonals bisect each other in a rectangle,rhombus, parallelogram, and a square (which is a type of rhombus).