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In the figure below, 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


See solution below
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