Question 847014
<pre>
{{{drawing(400,400,-1.5,1.5,-1.5,1.5,

line(-1,-1,1,-1),
line(1,-1,1,1),
line(1,1,-1,1),
line(-1,1,-1,-1),
line(-1,1,0,-1),
line(0,1,1,-1),
line(-1,0,1,1),
line(-1,-1,1,0),
locate(-1,-1,D),
locate(1,-1,C),
locate(0,-1,H),
locate(1.05,.05,G),
locate(-.21,-.43,I),
locate(.5,-.1,J) )}}}

Use the Pythagorean theorem on &#916;DCG

DG² = DC² + CG²
DG² = 1² + {{{(1/2)^2}}}
DG² = 1 + {{{1/4}}}
DG² = {{{5/4}}}
DG = {{{sqrt(5)/2}}}

&#916;DIH &#8765; &#916;DCG, therefore

{{{ID/(DH)}}}{{{""=""}}}{{{DC/DG}}}

ID×DG = DC×DH

ID{{{""*expr(sqrt(5)/2)}}}{{{""=""}}}{{{1*expr(1/2)}}}

Multiply through by 2

ID{{{sqrt(5)}}}{{{""=""}}}{{{1}}}

ID = {{{1/sqrt(5)}}} 

Rationalize the denominator:

ID = {{{1/sqrt(5)}}}{{{""*""}}}{{{sqrt(5)/sqrt(5)}}}

{{{sqrt(5)/5}}} 

&#916;CJD &#8765; &#916;DCG, therefore,

{{{JG/(CG)}}}{{{""=""}}}{{{CG/DG}}}

JG×DG = CG×CG

JG{{{""*expr(sqrt(5)/2)}}}{{{""=""}}}{{{expr(1/2)*expr(1/2)}}}

JG{{{""*sqrt(5)/2}}}{{{""=""}}}{{{expr(1/4)}}} 

Multiply through by 4

2JG{{{sqrt(5)}}}{{{""=""}}}{{{1}}}

JG = {{{1/(2*sqrt(5))}}} = {{{sqrt(5)/10}}}  (rationalizing)

DI + IJ + JG = DG

{{{sqrt(5)/5}}} + IJ + {{{sqrt(5)/10}}} = {{{sqrt(5)/2}}}

Multiply through by 10

{{{2*sqrt(5)}}} + 10·IJ + {{{sqrt(5)}}} = {{{5*sqrt(5)}}}

{{{3*sqrt(5)}}} + 10·IJ = {{{5*sqrt(5)}}}

                  10·IJ = {{{2*sqrt(5)}}}

                     IJ = {{{2*sqrt(5)/10}}}

                     IJ = {{{sqrt(5)/5}}}

So the area of the inner square is IJ² = {{{(sqrt(5)/5)^2}}} = {{{5/25}}} = {{{1/5}}} 
 
Edwin</pre>