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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:
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AE = CD
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The measure of angle DOE is 120º
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FH = GH
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The measure of angle FHG is 120º
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mÐHFG = mÐHGF
= 30º
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OB is the bisector of ÐDOE
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mÐOBC = mÐOEC
= mÐHGC
mÐADO = mÐABO
= mÐAFH
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OD is the bisector of ÐAOB
OE is the bisector of ÐBOC
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mÐAOB = mÐBOC
= 120º
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mÐCMG = mÐANF
= 30º
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OB and FG are perpendicular
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Triangle OPQ is equilateral
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Triangle JPH is equilateral
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Triangle HKQ is equilateral
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HJ = HK + OP
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See solution below
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