Question 1070409
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AB parallel to DE so CA is a transversal of two parallel lines.  Hence angle CDE congruent to angle CAB and angle CED congruent to angle CBA.  Angle C is congruent to itself.  So by AAA, triangle DEC is similar to triangle ABC.


Since AD is congruent to DC, D must be the midpoint of AC.  Therefore AC is twice the measure of AD or DC.  Therefore the sides of triangle DEC and the sides of triangle ABC are in proportion 1:2.


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

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