Area of square inside a triangle

Question

Considering the attached image, please compute the area of the square.

$\\$To elaborate suppose we have a triangle $\Delta ABC$, that its angle $\angle ABC$ is equal to $45^o$. On side $\overline{AB}$ we have a point $E$ that divide the side to two segments, $\overline{BE}=4$ and $\overline{EA}=6$.

Find the area of the square $\square CDEF$, such that its $\overline{FC}$ side reside on $\overline{AC}$ side of the triangle $\Delta ABC$, its one vertex is on E point, and the alternate vertex of E is on vertex C.

Answer 1

$$\begin{align*} \angle ACE &= 45^\circ\\ \triangle ABC &\sim \triangle ACE\\ \frac{AC}{AB} &= \frac{AE}{AC}\\ {AC}^2 &= 60 \end{align*} $$

Consider $\triangle ACE$, taking sine law and considering only the case where $\angle AEC = \angle ACB$ are obtuse,

$$\begin{align*} \frac{\sin\angle AEC}{AC} &= \frac{\sin \angle ACE}{AE}\\ \sin\angle AEC &= \frac{AC\sin\angle ACE}{AE}\\ &= \frac{\sqrt {60}\cdot \frac1{\sqrt2}}{6} = \frac{\sqrt{30}}{6}\\ \cos\angle AEC &= \color{red}-\sqrt{1-\sin^2\angle AEC}\\ &= \color{red}-\frac{\sqrt6}{6} \end{align*}$$

The area of the square can be found from side $EF = AE\cos \angle AEF$:

$$\begin{align*} (AE\cos \angle AEF)^2 &= {AE}^2 \cos^2(\angle AEC-45^\circ)\\ &= {AE}^2\cdot \frac{1+\cos 2(\angle AEC-45^\circ)}{2}\\ &= {AE}^2\cdot \frac{1+\cos (2\angle AEC-90^\circ)}{2}\\ &= {AE}^2\cdot \frac{1+\sin 2\angle AEC}{2}\\ &=\frac{6^2}2\left[1+2\cdot\frac{\sqrt{30}}{6}\left(\color{red}-\frac{\sqrt6}{6}\right)\right]\\ &= 18 \color{red}- 6\sqrt5 \end{align*}$$

If otherwise consider an acute $\angle ACB$ and flip the red $\color{red}-$ sign, this gives another solution.

Answer 2

Hint: In $\triangle BCE$, $D$ is the circumcenter. Can you see why?

So $BD = CD = DE$ and $\angle BED = \angle A$.

So you can easily see that $\sin 2A = \frac{2}{3}$ and area of the square is $36 \sin^2A = 18(1-\cos 2A)$

Answer 3

Hint: See picture. Make sure you understand how each sides are calculated and make use of similar triangles.

Answer 4

Similar to other answers. Identify $D$ as circumcenter of $\triangle BEC$. With more angle chasing, $AC$ is tangent to the circumcircle of $\triangle BEC$.

By secant-tangent theorem $AC^2=AE\cdot AB=60$

Denote side of square by $x$. Then by Pythagoras in $\triangle EAF$,

$$x^2 + (\sqrt{60}-x)^2=6^2$$

There'll be two values of $x$. Two squares are possible as explained by @peterwhy in the comment below the question. Obtain $x^2$.

Answer 5

Let the square have side $L$ and consider triangles $AEF$ and $BCE$.

Simply using the sum of angles in a triangle we see that $\angle AEF =\angle BCE=X$ say. Then $$6\sin X=\sqrt{36-L^2}, \sqrt2L\sin X=4\sin 45.$$ Eliminating $\sin X$ gives a quadratic for the area $L^2$, $$L^2(36-L^2)=12^2,$$ with solutions $L^2=18\pm 6\sqrt 5.$