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Geometry Problem 50 with Solution: Triangle with Equilateral triangles, (17) Conclusions

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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:

  1. AE = CD

  2. The measure of angle DOE is 120º

  3. FH = GH

  4. The measure of angle FHG is 120º

  5. mÐHFG = mÐHGF = 30º

  6. OB is the bisector of ÐDOE

  7. mÐOBC = mÐOEC = mÐHGC
    mÐADO = mÐABO = mÐAFH

  8. OD is the bisector of ÐAOB
    OE is the bisector of ÐBOC

  9. mÐAOB = mÐBOC = 120º

  10. mÐCMG = mÐANF = 30º

  11. OB and FG are perpendicular

  12. Triangle OPQ is equilateral

  13. Triangle JPH is equilateral

  14. Triangle HKQ is equilateral

  15. HJ = HK + OP

Equilateral triangle, 30, 60, 120 degrees
 

See solution below

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