Question 1210474
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In rectangle ABCD, corner A is folded over crease DE to point F on BC. Find BC.
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<pre>
Make a sketch following my descriptions/instructions.


We consider rectangle ABCD.  For simplicity, assume that the rectangle is "vertical",
i.e. its vertical dimension 'b' = AD = BC is greater than (or equal to) its horizontal dimension 'a' = AB = CD.


We fold corner A, so point A comes to point F on BC: so, point F is the image of point A
under folding, and F is on BC between points B and C.   <<<---===  It is in accordance with the given part.


Then it is clear that at this folding the side AB transforms to interval BF;
so, AB = BF = a.


It is also clear that at this folding the side AD becomes FD;  therefore, DF = b.


Consider right-angled triangle FCD. For its leg FC we can write

    FC = {{{sqrt(DF^2 - CD^2)}}} = {{{sqrt(b^2 - a^2)}}}.


Now interval BC is the sum of two intervals

    BC = BF + FC = a + {{{sqrt(b^2-a^2)}}}.


But BC = b; so, we can rewrite the last equation in the form

    b = a + {{{sqrt(b^2-a^2)}}}.    (1)


Now we will perform formal transformations with this equation.  From (1) we have, moving 'a' on the left side

    b - a = {{{sqrt(b^2-a^2)}}}.


Square both sides

    (b-a)^2 = b^2 - a^2,

    b^2 - 2ab + a^2 = b^2 - a^2


Cancel b^2 in both sides and continue     

    -2ab + a^2 = - a^2,

     a^2 + a^2 = 2ab,

     2a^2 = 2ab.


Cancel the common factor 2a in both sides

    a = b.


So, we proved that under given conditions, it must be a = b.


It is what the problem wants to be proved.


<U>ANSWER</U>.  Under given conditions, the rectangle ABCD is the square and its sides are congruent: a = b.
</pre>

Solved.


It is what the Artificial Intelligence (= @CPhill) missed and could not to prove properly.