Question 931693
<pre>
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green(arc(5,0,3,-3,180-21,180),arc(-3,3,3,-3,360-21,360)),
locate(-5,0,B),locate(5,0,C),locate(-3,3.5,A), locate(3,3.5,D),locate(0,1.8,E),
line(-3,3,5,0),line(-5,0,3,3) )}}}



Given:    Trapezoid ABCD,  
          AC = DB
To prove: AB = DC


AD &#8741; BC                          definition of a trapezoid
&#8736;DAC = &#8736;BCA, &#8736;ADB = &#8736;DBC       alternate interior angles
&#916;ADE &#8765; &#916;CBE                     two angles equal in each  
AE/EC = DE/EB                   corresponding sides of similar 
                                        triangles are proportional
AE/EC+1 = DE/EB+1               adding the same quantity, 1, to both sides
AE/EC+EC/EC = DE/EB+EB/EB       replacing 1 by EC/EC and by EB/EB
(AE+EC)/EC = (DE+EB)/EB         adding fractions 
AE+EC = AC,  DE+EB = DB         a whole is the sum of its parts           
AC/EC = DB/EB                   Substitution of equals for equals
AC = DB                         Given
AC/EC = AC/EB                   Substitution of equals for equals
AC*EB = AC*EC                   Cross-multiplication or product of
                                extremes equal product of means.
EB = EC                         Dividing both sides by AC                             
&#916;EBC is isosceles               two sides equal
&#8736;ECB = &#8736;EBC                    base angles of an isosceles triangle
AC = DB                         given
BC = BC                         identity
&#8736;ACB = &#8736;DBC                    Same as &#8736;ECB = &#8736;EBC 
&#916;ABC &#8773; &#916;DCB                    SAS   
AB = DC                         Corresponding parts of congruent triangles

Edwin</pre>