A proof related to geometric mean. Can anyone solve this problem??


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*If a square is inscribed into a right triangle in such a way that one side of the square lies on the hypotenuse, then this side is the geometric mean between the two remaining segments of the hypotenuse.


The definition of the geometric mean in my text book is this.


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*The geometric mean between two segments $a$ and $c$ is the third segment $c$ such that $a:c=c:b.$


I drew this figure.
So I tried to show that the angle ACE is right. Then, there is theorem that says $CD$ is geometric mean of $AD$ and $DE$ in $\triangle ACE.$ Then, I will get $CD^2 = AD*DE$ and $FE^2 = DE*EA'.$ If I multiply this two, then I get $DE^4 = AD*EA'*DE^2 \Rightarrow DE^2=AD*EA'$ because $CD=FE=DE.$ But I cannot show that $\angle ACE=90^{\circ}.$ Can anyone help me with this or give other proof?? thanks.
 A: We use your picture (thanks, it saves a lot of words). 
Triangles $ACD$ and $FA'E$ are similar, because each is similar to the big triangle. Note that in the similarity, $AD$ and $FE$ are corresponding sides. 
Thus if $x$ is the side of the square, then
$$\frac{x}{AD}=\frac{A'E}{x},$$
and we are finished. 
Remark: The angle $ACE$ is a right angle only if our big triangle is isosceles. Thus the strategy proposed in the post will not work in general.
A better way of visualizing the answer above is to remove everything from the picture except the two small triangles on the left and right, and slide $\triangle FEA'$ leftward until the two triangles meet. We end up with a triangle $ACA'$, right-angled at $C$. The altitude $CD$ divides $\triangle ACA'$ into two similar triangles. 
A: Taking the side of the square as $s$, note that $s\cot A=AD$ and $s\cot A'=A'E$ and $\cot A\cdot \cot A'=\cot A\tan A=1$. Thus, $AD\cdot A'E=s\cot A\cdot s\cot A'=s^2$, which proves the proposition. 
You can also do this without invoking trigonometric functions, since all we basically used here was similarity.
