Question 174284
Given: ABCD is a square and XYZW is a figure inside the square
AX is congruent to BY which is congruent to CZ which is congruent to DW
BX is congruent to CY which is congruent to DZ which is congruent to AW 
Prove that WXYZ is a square
<pre><font size = 4 color = "indigo"><b>
I'll just outline the proof.  You will have to write it
up as a two-column proof.

{{{drawing(400,400,-3,3,-3,3,
 
rectangle(-2,-2,2,2), 
 
triangle(-2,-2,-1,1,-2,2),
triangle(-2,-2,-1,1,-1,-1),
triangle(-2,-2,-1,-1,2,-2),
triangle(2,-2,1,-1,-1,-1),
 
triangle(1,-1,2,2,1,1),
triangle(-2,2,2,2,1,1),
triangle(-2,2,-1,1,1,1),
locate(2,2.3,C),
locate(-2,-2,A),locate(2,-2,B), locate(-1,-1,X),
locate(1,-1,Y), locate(1,1,Z), locate(-1,1,W),
locate(-2,2.3,D))}}}


Now these four outer triangles shown below
are easily proved congruent by SSS:

{{{drawing(400,400,-3,3,-3,3,
triangle(-2,2,2,2,1,1),
triangle(2,2,2,-2,1,-1),
triangle(-2,-2,2,-2,-1,-1),
triangle(-2,-2,-2,2,-1,1),
locate(2,2.3,C),
locate(-2,-2,A),locate(2,-2,B), locate(-1,-1,X),
locate(1,-1,Y), locate(1,1,Z), locate(-1,1,W),
locate(-2,2.3,D))}}}

Now by subtracting the sum of the measures of a
pair of congruent angles in each corner of the 
big given square from 90° you get that these 
four angles at A, B, C, and D, shown below are 
all congruent:

{{{drawing(400,400,-3,3,-3,3,
line(2,2,1,-1),
line(2,2,1,1),
line(-2,2,1,1),
line(-2,2,-1,1),
line(-2,-2,-1,1),
line(-2,-2,-1,-1),
line(2,-2,-1,-1),
line(2,-2,1,-1),
locate(2,2.3,C),
locate(-2,-2,A),locate(2,-2,B), locate(-1,-1,X),
locate(1,-1,Y), locate(1,1,Z), locate(-1,1,W),
locate(-2,2.3,D))}}}

Therefore the four triangles shown below around 
the center quadrilateral are congruent by SAS:

{{{drawing(400,400,-3,3,-3,3,
triangle(-2,2,-1,1,1,1),
triangle(2,2,1,1,1,-1),
triangle(2,-2,1,-1,-1,-1),
triangle(-2,-2,-1,-1,-1,1),
locate(2,2.3,C),
locate(-2,-2,A),locate(2,-2,B), locate(-1,-1,X),
locate(1,-1,Y), locate(1,1,Z), locate(-1,1,W),
locate(-2,2.3,D))}}} 

Therfore the center figure WXYZ is at least a 
rhombus, that is, all four sides are congruent,
by "corresponding parts of congruent triangles".

Now let's go back to the original figure:

{{{drawing(400,400,-3,3,-3,3,
rectangle(-2,-2,2,2), 
triangle(-2,-2,-1,1,-2,2),
triangle(-2,-2,-1,1,-1,-1),
triangle(-2,-2,-1,-1,2,-2),
triangle(2,-2,1,-1,-1,-1),
triangle(1,-1,2,2,1,1),
triangle(-2,2,2,2,1,1),
triangle(-2,2,-1,1,1,1),
locate(2,2.3,C),
locate(-2,-2,A),locate(2,-2,B), locate(-1,-1,X),
locate(1,-1,Y), locate(1,1,Z), locate(-1,1,W),
locate(-2,2.3,D))}}}

Finally at each corner point of the inside figure,
X, Y, Z, and W, there are 4 adjacent angles which have
sum 360°.  Three of the angles at each of those
corner points X,Y,Z, and W, are congruent to three 
of the angles at each of the other corner point.

So by subtracting the measures of those three angles 
from 360°, we get that each of the interior angles of the 
the inner quadrilateral WXYZ have the same measure.

Now since the sum of the measures of the interior
angles of any quadrilateral is (4-2)×180° or 360°,
each of the corner points of the inside quadrilateral
must be 360°÷4 or 90°.  Thus it is a rhombus with
four 90° interior angles, which is a square.   

Edwin</pre>