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<td><mdo:image alt="Triangle with Equalateral Triangles on sides" src="napoleon.gif"></mdo:image></td>
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<p>Triangle ABC has equilateral triangles drawn on its edges.
Points P, Q and R are the centres of the equilateral triangles.
What can you prove about the triangle PQR?</p>
<p>You can change triangle ABC below by dragging the vertices and
observe what happens to triangle PQR.</p>
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<p><mdo:applet height="345" width="593" code="GSP.class" archive="JSPDR3.jar" datafile=""><param name="MeasureInDegrees" value="1" ></param><param name="DirectedAngles" value="0" ></param><param name="BackRed" value="255" ></param><param name="BackGreen" value="255" ></param><param name="BackBlue" value="255" ></param><param name="Construction" value="{1} Point(67,135)[label('A'),blue]; {2} Point(120,63)[label('B'),blue]; {3} Point(211,134)[label('C'),blue]; {4} Segment(2,1)[black]; {5} Segment(3,2)[black]; {6} Segment(1,3)[black]; {7} Circle(1,2)[hidden,black]; {8} Circle(2,1)[hidden,black]; {9} Circle(2,3)[hidden,black]; {10} Circle(3,2)[hidden,black]; {11} Circle(3,1)[hidden,black]; {12} Circle(1,3)[hidden,black]; {13} Intersect1(7,8); {14} Intersect1(8,7)[hidden]; {15} Intersect1(9,10); {16} Intersect1(10,9)[hidden]; {17} Intersect1(11,12); {18} Intersect1(12,11)[hidden]; {19} Segment(13,1)[black]; {20} Segment(2,13)[black]; {21} Circle(13,2)[hidden,black]; {22} Segment(14,13)[hidden,black]; {23} Segment(15,2)[black]; {24} Segment(3,15)[black]; {25} Circle(15,3)[hidden,black]; {26} Segment(16,15)[hidden,black]; {27} Segment(17,3)[black]; {28} Segment(1,17)[black]; {29} Circle(17,1)[hidden,black]; {30} Segment(18,17)[hidden,black]; {31} Intersect2(8,21)[hidden]; {32} Intersect2(10,25)[hidden]; {33} Intersect2(12,29)[hidden]; {34} Segment(31,1)[hidden,black]; {35} Segment(32,2)[hidden,black]; {36} Segment(33,3)[hidden,black]; {37} Intersect(34,22)[label('P')]; {38} Intersect(35,26)[label('Q')]; {39} Intersect(30,36)[label('R')]; {40} Segment(39,37)[red]; {41} Segment(38,39)[red]; {42} Segment(37,38)[red];" ></param></mdo:applet></p>
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NOTES AND BACKGROUND<br></br>
<p>There are many ways of proving this result. One way you might like to try involves tessellation.</p> 
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(1) Draw any triangle, with angles $A, B$ and $C$ say.\par
(2) Draw equilateral triangles $T_1, T_2$ and $T_3$ on the three sides of 
$\Delta ABC$.\par
(3) Fit copies of the original triangle and $T_1, T_2$ and $T_3$ into a 
tessellation pattern so that, at each vertex of the tessellation, the angles 
are $A, B$ and $C$ and three angles of $60^o$ making an angle sum of $360^o$.\par
(4) Napoleon's Theorem can be proved by simple geometry using a small part of 
this pattern without even assuming that this tessellation extends indefinitely 
in all directions, which is intuitively obvious but requires advanced 
mathematics to prove it. \par
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<mdo:flash height="400" width="400"><param value="/content/98/12/15plus5/triangles2.swf" name="movie" ></param><param value="8" name="flashplayerversion" ></param></mdo:flash> <br></br>
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Van Aubel's Theorem is related to Napoleon's Theorem. Van Aubel's Theorem 
states that if four squares are drawn on the edges of {\bf any} quadrilateral then 
the lines joining the centres of the squares on opposite edges are equal in 
length and perpendicular.
  \par
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<p> For an animated proof of Van Aubel's Theorem see <a href="http://agutie.homestead.com/files/vanaubel.html">http://agutie.homestead.com/files/vanaubel.html</a></p></div>
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