Question 1130860
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proof: the diagonals are perpendicular to each other



 assuming we have a rhombus {{{ABCD}}} with diagonals {{{AC}}} and {{{BD}}} intersecting at point {{{E}}}


{{{AE}}} &#8773; {{{CE}}}..................as two halves of a diagonal {{{AC}}}


{{{BE}}} &#8773; {{{BE}}}................. by reflexive property of congruence, a segment is congruent to itself 

{{{AB}}} &#8773; {{{CB}}}.....................as sides of a rhombus

&Delta; {{{ AEB }}} &#8773; &Delta; {{{CEB}}}..........by SSS congruence
< {{{AEB}}} &#8773; < {{{CEB}}}....................... corresponding angles
< {{{AEB}}} + < {{{CEB}}}={{{180}}}°.........since they are supplementary angles
< {{{AEB}}} = < {{{CEB}}}={{{90}}}°..........addition property of congruent angles

{{{AC}}} &perp; {{{BD}}}..........the diagonals of {{{ABCD}}} are perpendicular to each other