🚀 go-pugleaf

Miro Jurisic wrote:
> 
> Now that we are talking about cyclic quadrilaterals...
> 
> Give a Euclidean (one-edge ruler + compass) construction of a cyclic
> quadrilateral given its four sides.
> 
> meeroh
> 
> --
> Miroslav Jurisic | meeroh@mit.edu | http://www.mit.edu/people/meeroh/
> MIT Information Systems (Mac development) | Athena On-Line Consulting
> Lt. Commander Meeroh Ambellus, ATS TrekMUSH  (ats.trekmush.org:1701)
> 
> "It's not hatred that's important; it's the desire to anihilate." [101
> Dalmatians]

Solution:

   For a quad.ABCD,                A--------B
                                  /          \
                                 D------------C
   I try to split into 2 triangles ABD and BCD,
   then angle A + angle C = 180,
                       BD = BD


   Use a straight edge and fix the length of DA and CB with A,C
coincide,

                         C'
   ie.             D-----A--------B'

   Use a pair of compasses and draw the perpendicular bisector of DB',
   Use a pair of compasses and use A as centre, draw an arc with AB as
radius.
   Join the point when the arc cuts the perpendicular bisector.

                           X  
                           |
                           |
                   D-----A--------B'

   The point fixed is actually the location of B of the cyclic
quadrilateral.

   Now use D as centre, draw an arc DC.
       Use X as centre, draw an arc BC.
   
   Join all the lines, the cyclic quadrilateral is constructued.