Question 513348
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<b>Theorem</b> If ABCD is a <A HREF=http://www.algebra.com/algebra/homework/Rectangles/Different-between-parallelogram-rectangle-square-rhombus-and-trapezoid.lesson>parallelogram</A>, then prove that the diagonals of ABCD bisect each other.



{{{drawing( 160, 160, -10, 10, -10, 10, line( -6, -6,4,-6) , line( -6, -6,-2,2), line( -2, 2,8,2) , line( 8, 2,4,-6),line( -6, -6,8,2) ,line( -2, 2,4,-6) ,locate( -6.5,-6.5,A),locate( 4.5,-6.5,B),locate( 8.5,3.5,C),locate(-2.5,4,D),locate(-0,-2.4,O))}}}



<b>Proof</b>


Let the two diagonals be AC and BD and O be the intersection point.


We have to prove that O is the midpoint of AC and also the midpoint of BD.


Hence, {{{AO=OC}}} and {{{BO=OD}}}


We will prove using <A HREF=http://www.algebra.com/algebra/homework/Triangles/Geometry-Proving-Triangles-Congruent.lesson>congruent triangles </A>concept.


Consider two <A HREF=http://www.algebra.com/algebra/homework/Triangles/Triangles-and-its-basic-properties.lesson>Triangles </A> ABO and COD.


1. {{{ angle OAB = angle CAB = angle ACD = angle OCD}}}   ....( Line AC is a transversal of the parallel lines AB and CD, hence alternate angles).


2. {{{angle ODC = angle BDC = angle DBA = angle OBA}}}   ....(Line BD is a transversal of the parallel lines AB and CD, hence alternate angles).


3. {{{angle DOC = angle AOB}}}                           ....(Opposite angles when two lines intersect each other area equal)


From conditions 1,2 and 3


Triangle ABO is similar to triangle CDO (By <A HREF=http://www.algebra.com/algebra/homework/Triangles/Angle.wikipedia>Angle </A>-Angle similar property)


Since Triangles are similar, Hence ratio of sides are equal from <A HREF=http://www.algebra.com/algebra/homework/Triangles/Geometry-Similar-Triangles.lesson>similar triangles </A>property.


{{{(DC/AB)=(DO/OB)=(CO/OA)}}}     .........(4)


From theorem that <A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Opposite-sides-of-a-parallelogram-are-equal.lesson>Opposite sides of a parallelogram are equal</A>,


{{{DC=AB}}}                       ..........(5)


From equation (4) and (5)


{{{(DC/AB)=(DO/OB)=(CO/OA)=1}}}


{{{DO/OB = 1}}}


{{{DO = OB}}}


Similarly, {{{CO=OA}}} 


Hence, We conclude that AO = CO and BO = DO.


QED


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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