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 Problem 50. Triangle with Equilateral triangles. Level: High School, SAT Prep, College

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 120

  3. FH = GH

  4. The measure of angle FHG is 120

  5. mHFG = mHGF = 30

  6. OB is the bisector of DOE

  7. mOBC = mOEC = mHGC
    mADO = mABO = mAFH

  8. OD is the bisector of AOB
    OE is the bisector of BOC

  9. mAOB = mBOC = 120

  10. mCMG = mANF = 30

  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
 

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