What is the value of $x$ in $ABCD$ rectangle where $AE = 4, BE = 6, CE=5$ and $DE = x$?

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

I can not relate these information with $$DE$$.

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

Another way is:

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

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

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

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

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 $$x=\sqrt5.$$

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.