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Problem 50. Triangle with Equilateral triangles. Level: High School, SAT Prep, College
 In the figure above, equilateral triangles ABD and BCE are drawn on the sides of a triangle ABC. F, G, and H are the midpoints of AD, CE, and AC respectively. HL and FG are perpendicular. Lines FH, CD, HL,  and FG are cut by line AE at J, O, P, and N respectively. Lines GH, HL, and FG are cut by line CD at K, Q, and M respectively. Prove the following: AE = CD The measure of angle DOE is 120º FH = GH The measure of angle FHG is 120º mÐHFG = mÐHGF = 30º OB is the bisector of ÐDOE mÐOBC = mÐOEC = mÐHGC mÐADO = mÐABO = mÐAFH OD is the bisector of ÐAOB OE is the bisector of ÐBOC mÐAOB = mÐBOC = 120º mÐCMG = mÐANF = 30º OB and FG are perpendicular Triangle OPQ is equilateral Triangle JPH is equilateral Triangle HKQ is equilateral HJ = HK + OP Using the deductive method, we start with a few true statements (the axioms) and use them to prove dozens, hundreds, or thousands,  of other statements (the theorems). Geometry was organized by the Greek mathematician Euclid, and his deductive method is still used by mathematicians today.   Hints: See Geometry Help
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