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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:

  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

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.

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Last updated: January 24, 2008

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