Question 899843
1.

{{{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.

 Lesson (Proof: The diagonals of parallelogram bisect each other) was created by tutor chillaks.

2.
what are the lengths of the segments into which the y-axis divides the segment joining ({{{-6}}},{{{-6}}}) and ({{{3}}},{{{6}}}) ? 
 
{{{y-axis}}} means {{{x= 0}}} 
 
Let the ratio be {{{k}}}:1 
 The {{{0= ((-6)*1+3*k)/(k+1)}}}
 
 => {{{0= (-6+3k)/(k+1)}}}
 
 => {{{-6+3k=0}}} =>;{{{k=2}}}
 
 Hence {{{y= (-6*1+6*2)/(2+1) =(-6+12)/3 =2 }}}

 Hence intersection point is C ({{{0}}},{{{2}}})
 
 Hence Length A ({{{-6}}},{{{-6}}}) to C ({{{0}}},{{{2}}}) ={{{sqrt((0-(-6))^2+(2-(-6))^2))=sqrt(6^2+8^2) =sqrt(36+64)=sqrt(100) =10}}} units
 
 Length C  ({{{0}}},{{{2}}}) to B ({{{3}}},{{{6}}}) ={{{sqrt((3-0)^2+(6-2)^2)=sqrt((3)^2+(4)^2) =sqrt(9+16)=sqrt(25) =5}}} units