Question 1113279
<pre>
The answer to problem as stated is no.  However there is a
similar problem to which the answer is yes. First the problem
as stated is no:

This figure below is drawn to scale. The longer side of the
large rectangle is {{{phi=((1+sqrt(5))/2)}}} times the shorter
side.

{{{drawing(400,2000/7,-.5,3.7,-.5,2.5,

line(0,0,1+sqrt(5),0),
line(1+sqrt(5),0,1+sqrt(5),2),
line(0,2,1+sqrt(5),2),
line(0,0,0,2), circle(1,1,1), line(1,0,1,2)  )}}}  

All golden rectangles are similar.  So the answer is 
no, because neither of those two rectangles is similar
to the large rectangle.

-------------------------------

Now here is a similar problem where the answer is yes.
We draw the line different: 
</pre>
Take a Golden Rectangle and draw the largest circle
inside it that touches three sides. If we draw a line
parallel to the shorter side and tangent to the circle
and then cut the rectangle into two pieces along that 
line, will either of the two smaller rectangles be a 
Golden Rectangle?
<pre>
The big rectangle below is Golden.  The short side is "a"
and the long side is {{{a*phi}}}.  The circle is inscribed
in a square. The bottom side of the square is "a" units
So the short side of the small rectangle on the right is 
{{{a*phi-a}}} 

{{{drawing(400,2000/7,-.5,3.7,-.25,2.75,
locate(3.3,1,a),locate(2.5,0,a*phi-a),


line(0,0,1+sqrt(5),0),
line(1+sqrt(5),0,1+sqrt(5),2),
line(0,2,1+sqrt(5),2),
line(0,0,0,2), circle(1,1,1), line(2,0,2,2)  )}}} 
      |------------------------------------------|
                          {{{a*phi}}}

Now we will show that the small rectangle on the right is golden.

The ratio of the sides is

{{{a/(a*phi-a)}}}

Divide numerator and denominator by a

{{{1/(phi-1)}}}

Substitute {{{phi=((1+sqrt(5))/2)}}}

{{{1^""/((1+sqrt(5))/2-1)}}}

Multiply top and bottom by 2

{{{2/((1+sqrt(5))-2)}}}

{{{2/(1+sqrt(5)-2)}}}

{{{2/(-1+sqrt(5))}}}

Rationalize the denominator by multiplying by {{{(-1-sqrt(5))/(-1-sqrt(5))}}}

{{{2/(-1+sqrt(5))}}}{{{""*""}}}{{{(-1-sqrt(5))/(-1-sqrt(5))}}}

{{{(-2-2sqrt(5))/(1-5))}}}

{{{(-2-2sqrt(5))/(-4))}}}

{{{(-2(1+sqrt(5)))/(-4))}}}

Divide top and bottom by -2:

{{{(1+sqrt(5))/2)}}}

Which is {{{phi}}}, the golden ratio.

So for this problem, the answer is 

Yes, the small rectangle on the right is golden.

[I have a hunch this was the problem that was intended.
Point this out to your teacher.]

Edwin</pre>