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Tiffany N. KeysŐ Essay 2:

NapoleonŐs Triangle

INVESTIGATION:  Given any triangle ABC, construct equilateral triangles on each side and find the center of each equilateral triangle. The triangle formed by these three centers is Napoleon's Triangle.

Step 1:     Given triangle ABC, construct the point D such that ABD is an equilateral triangle, and D and C are on opposite sides of AB.

Step 2:     Construct the point E such that BCE is equilateral.

Step 3:     Construct the point F such that CAF is equilateral.

Step 4:     Let J be the center of ABD, K the center of BCE, and L the center of CAF.

Step 5:     Then JKL is Napoleon's triangle.

PROOF

Step 1:     Construct the point O such that triangles DAO and ABC are congruent.

Step 2:     Construct the point N such that BDN and ABC are congruent.

Step 3:     Construct the Napoleon triangles JMI and JGH from triangles DAO and BDN.

Step 4:     Considering triangle TEB, it is assumed that BE = BC and BT = BN = AC.

Step 5:     Since angle DBN is equal to angle BAC, it is assumed that <EBT = 180 - <ABC - <BAC = <ACB.

This shows that triangle TEB must be congruent to triangle ABC by SAS.

Step 6;     By construction, GK = KL, which also shows that GK = KL = LM = MI = IH = HG.

Step 7:     JL = JG by construction, GK = LK by STEP 6, and JK = JK, so that triangles GJK and LJK are congruent by SSS.

Step 8:     Therefore we can assume <KJG = <LJK and that the six angles at J must all be congruent and all are 60ˇ.

Step 9:     Since <LJK = 60 degrees, we can assume that <JKL and <KLJ are 60ˇ.

Step 10:   In conclusion, triangle JKL is equilateral.