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.

(In reply to re: Trying to put a picture into words... by WryTangle)

I guess I got a bit carried away with my pretty picture:

<o:p> </o:p>

Take two…

<o:p> </o:p>

In our nonagon AC = BD, so just draw the two parallel diagonals BD & AE. 

Now we have:

<v:shapetype id=_x0000_t75 stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" o:preferrelative="t" o:spt="75" coordsize="21600,21600"><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path o:connecttype="rect" gradientshapeok="t" o:extrusionok="f"></v:path><o:lock aspectratio="t" v:ext="edit"></o:lock></v:shapetype><v:shape id=_x0000_s1026 style="MARGIN-TOP: 3.35pt; Z-INDEX: 1; MARGIN-LEFT: 236.35pt; WIDTH: 180pt; POSITION: absolute; HEIGHT: 175pt; mso-position-horizontal: absolute; mso-position-horizontal-relative: text; mso-position-vertical: absolute; mso-position-vertical-relative: text" type="#_x0000_t75"><v:imagedata o:title="" src="file:///C:DOCUME~1AlecLOCALS~1Tempmsohtml11clip_image001.emz"></v:imagedata><w:wrap type="square"></w:wrap></v:shape>ABC = BCD = 140.

CBD = CDB = 20.

<v:shapetype id=_x0000_t202 path="m,l,21600r21600,l21600,xe" o:spt="202" coordsize="21600,21600"><v:stroke joinstyle="miter"></v:stroke><v:path o:connecttype="rect" gradientshapeok="t"></v:path></v:shapetype><v:shape id=_x0000_s1027 style="MARGIN-TOP: 53.2pt; Z-INDEX: 2; MARGIN-LEFT: 289pt; WIDTH: 40.75pt; POSITION: absolute; HEIGHT: 27.15pt; mso-position-horizontal: absolute; mso-position-vertical: absolute" type="#_x0000_t202" stroked="f" filled="f" strokecolor="blue"></v:shape> <TABLE cellSpacing=0 cellPadding=0 width="100%"> <TBODY> <TR> <TD style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; BORDER-BOTTOM: #ece9d8; BACKGROUND-COLOR: transparent">

</TD></TR></TBODY></TABLE>Thus ABD = BDE = 120.

Hence BAE = DEA = 60.

Let P lie on AE such that DBP = APB = 60.

All angles = 60 says triangle ABP is equilateral therefore AB = AP.

BPE = 120 = BDE, and DBP = DEP = 60 Therefore BDEP is a parallelogram and BD = PE.

Therefore as AE = AP + PE, AE = AB + AC.

Posted by Alec on 2004-11-26 17:08:12