Question 1209431
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Let point E represent the intersection of diagonals AC and BD.
{{{
drawing(400,400,-4,5.4513,-3,3.3579,
line(-3,-2,2,-2),line(2,-2,4.4513,2.3579),line(4.4513,2.3579,-0.5487,2.3579),line(-0.5487,2.3579,-3,-2),line(-3,-2,4.4513,2.3579),line(-0.5487,2.3579,2,-2),

circle(-3,-2,0.06),circle(-3,-2,0.08),circle(-3,-2,0.1),circle(-3,-2,0.12),circle(2,-2,0.06),circle(2,-2,0.08),circle(2,-2,0.1),circle(2,-2,0.12),circle(4.4513,2.3579,0.06),circle(4.4513,2.3579,0.08),circle(4.4513,2.3579,0.1),circle(4.4513,2.3579,0.12),circle(-0.5487,2.3579,0.06),circle(-0.5487,2.3579,0.08),circle(-0.5487,2.3579,0.1),circle(-0.5487,2.3579,0.12),circle(0.7257,0.1789,0.06),circle(0.7257,0.1789,0.08),circle(0.7257,0.1789,0.1),circle(0.7257,0.1789,0.12),

arc(-3,-2,2,2,329.6789,360),
arc(-3,-2,2.5,2.5,299.3578,329.6789),
arc(-0.5487,2.3579,2,2,59.6789,119.3578),
arc(-0.5487,2.3579,2.5,2.5,0,59.6789),
line(-1.66,-1.62,-2.38,-1.86),
line(-1.9,-0.9,-2.44,-1.42),
line(-0.8,1.16,-0.7,1.74),line(-0.48,1.72,-0.46,1.22),
line(0.6,1.42,0.06,1.9),line(0.34,2.16,0.92,1.82),
line(1.86,2.3579,1.52,2.68),line(1.86,2.3579,1.5,2.1),line(-0.34,-2,-0.66,-1.76),line(-0.34,-2,-0.74,-2.32),

locate(-3.2,-2.2,"A"),locate(1.8,-2.2,"B"),locate(4.2513+0.3,2.1579+0.15,"C"),locate(-0.7487-0.3,2.1579+0.4,"D"),locate(0.5257,-0.0211,"E")
)
}}}
I'll provide a portion of the proof table.
<table border = "1" cellpadding = "5"><tr><td>Number</td><td>Statement</td><td>Reason</td></tr><tr><td>1</td><td>Segment AC bisects angle DAB</td><td>Given</td></tr><tr><td>2</td><td>Segment DB bisects angle ADC</td><td>Given</td></tr><tr><td>3</td><td>Segment AB is parallel to segment DC</td><td>Given</td></tr><tr><td>4</td><td>Angle DAC = Angle CAB</td><td>Definition of Angle Bisection</td></tr><tr><td>5</td><td>Angle ADB = Angle CDB</td><td>Definition of Angle Bisection</td></tr><tr><td>6</td><td>Angle CAB = Angle ACD</td><td>Alternate Interior Angles</td></tr><tr><td>7</td><td>Angle DAC = Angle ACD</td><td>Transitive Property</td></tr><tr><td>8</td><td>Segment DE = Segment DE</td><td>Reflexive Property</td></tr><tr><td>9</td><td>Triangle AED = Triangle CED</td><td>AAS Congruence Theorem</td></tr><tr><td>10</td><td>Segment AD = Segment DC</td><td>CPCTC</td></tr></table>
I'll let the student finish up from here.
The goal is to show that AB = BC = CD = AD.
So far the proof table above has demonstrated that AD = DC. 


Abbreviations:
AAS = angle angle side
CPCTC = corresponding parts of congruent triangles are congruent
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