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Nonagon (Posted on 2004-11-25) Difficulty: 4 of 5
Draw a regular nonagon. If we label the vertices of the polygon sequentially, starting from A, we can draw three line segments: AB, AC, and AE.

Show that AB + AC = AE.

See The Solution Submitted by SilverKnight    
Rating: 3.0000 (7 votes)

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Solution A simple geometric proof (no trig) | Comment 6 of 13 |
Draw line AC.
Draw line CE.
Since the measure in degrees of interior angles of a regular polygon is (n-2)(180)/n, angle B = 140.
Therefore, since AB = BC, triangle ABC is isosceles and angles BAC and BCA = 20.
Similarly with triangle CDE, angle DCE = 20.
Since triangle ABC is congruent to triangle CDE (by SAS), AC = CE.
Therefore triangle ACE is isosceles.
Since angle BCD = 140 and angles BCA and DCE each = 20, angle ACE = 100 and base angles CAE and CEA = 40.
Combining angles CAE and BAC, angle BAE = 60.
Draw line from B to point P on AE such that BP = AB. Label the intersection of AC and BP as point Q.
Since AB = BP, triangle BAP is isosceles, but since angle BAP = 60, triangle ABP is actually equilateral, and AP = AB.
Therefore, since AP + PE = AE, if we can show that PE = CE (which = AC), we will prove that AB + AC = AE.
In triangle APQ, since angle QAP = 40 and angle APB = 60, angle AQP = 80.
Therefore, vertical angle BQC = 80.
Since angle ABP = 60 and angle ABC =140, angle PBC = 80.
Since BP = BC and angle PBC = 80, triangle PBC is isosceles with base angles BPC and BCP = 50.
Since angle BCD =140, angle BCP = 50 and angle DCE = 20, angle PCE = 70.
Since angle APB = 60 and angle BPC = 50, angle CPE =70.
Since angles CPE and PCE = 70, triangle CEP is isosceles and CE = PE.
QED

  Posted by WryTangle on 2004-11-26 19:57:55
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