Question 802800
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I think that you are trying to say that E is the midpoint of AC and D is the midpoint of BC.  If this is the case, DE has to be parallel to AB, you can look up the theorem for yourself.  That means that angle BAC is congruent to angle DEC by one of the lemmas to the theorem for parallel lines and a transversal, and similarly angle ABC has to be congruent to angle EDC.  Then, since angle C is congruent to angle C by reflexive equality, triangle ABC is similar to triangle EDC.  The required proportions follow from the properties of similarity.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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