Given a square $ABCD$ with an internal point $E$, find the length of $CE$

Question

As title suggests, we have a square $ABCD$ and $E$ lies inside it. $AE=23$ and $BE=8$, the goal is to find the length $CE$:

This problem is fairly simply but quite interesting, there are many ways to solve this and I’m posting it here to see in what other ways it can be solved. I’ll post one possible solution, my own, as an answer below

Answer 1

Draw $EH\perp AB$ and $EH\perp BC$

then $AB=BC=\sqrt {593}, EH=KB=\frac{184}{\sqrt {593}}, HB=EK=\frac{64}{\sqrt {593}}$ and $CK=\frac{409}{\sqrt {593}}$

applying pytha in $\triangle EKC$

$EC^2= \frac{167281}{593} + \frac{4092}{593}=289$

$EC=17$

Trigonometric Solution:

$\sin\angle AEB = \cos \angle CBE = \frac{23}{\sqrt {593}} $

Applying Cosine Theorem

$EC^2=BC^2+EB^2-2\times BC \times EB \times \cos \angle CBE$

$EC^2 = 593 +64 - 2 \times \sqrt {593} \times 8 \times \frac{23}{\sqrt {593}}=289$

$EC=17$

OR

Using your approach

Draw EF

Since $\triangle EBF$ is an isoslceles right triangle, $BF=8$ , $EF=8\sqrt2 $ and $\angle EFC=45^{\circ}$

$EC^2=FC^2+EF^2-2\times FC \times EF \times \cos \angle EFC$

$EC^2=529+128-2\times 23 \times 8\sqrt2 \times \frac{1}{\sqrt2}$

$EC=17$

Answer 2

This is my approach:

Rotate $\triangle AEB$ by $90^\circ$ clockwise about $B$ such that $\triangle CFB$ is formed. Extend segment $AE$ such that it meets $CF$ at $H$. Notice that $EBFH$ is a square, therefore $EH=8$ and $CH=15$. Now by applying the Pythagorean on $\triangle CEH$ we get $CE=17$