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In the diagram of rectangular $ABCD$, $AE=4, BE= 6, CE = 5$ and $DE=x$, find the value of $x$

Source: Bangladesh Math Olympiad 2015 Junior Catagory

I can not relate these information with $DE$.

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    $\begingroup$ Hint: $AE^2 + CE^2 = BE^2 + DE^2$. You can verify this by making perpendicular foot of $E$ to each side of rectangular $ABCD$. $\endgroup$ – Doyun Nam Jan 31 at 3:44
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Another way is:

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$$\begin{cases}6^2=a^2+b^2\\ 5^2=b^2+c^2\\ x^2=c^2+d^2\\ 4^2=a^2+d^2\end{cases} \Rightarrow \\ x^2=(a^2+d^2)-(a^2+b^2)+(b^2+c^2)=\\ 4^2-6^2+5^2=5 \Rightarrow x=\sqrt{5}.$$

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$AE^2+CE^2=DE^2+BE^2$

This is called the $British$ $Flag$ $Theorem$.

https://en.wikipedia.org/wiki/British_flag_theorem

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Let $DAEF$ be parallelogram. Thus, $EBCF$ is also parallelogram,

which says $DF=AE=4$ and $FC=EF=6.$

We see that in the quadrilateral $DECF$ holds $DC\perp EF$, which says $$DE^2+FC^2=DF^2+EC^2$$ or $$x^2+6^2=4^2+5^2$$ or $$x^2=5,$$ which gives $x=\sqrt5.$

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Without loss of generality let $F$ be on $BC$ such that $EF\perp BC$ and $EF=x$. Then $FB=\sqrt{36-x^2}$. Extending $EF$ to meet $AD$ at $G$, $ AG=FB$ and so $EG=\sqrt{16-(36-x^2)}=\sqrt{x^2-20}$. Similarly, $FC=\sqrt{25-x^2}=GD$, so $DE=\sqrt{(x^2-20)+(25-x^2)}=\sqrt5$. All these manipulations use the Pythagorean theorem.

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