Solving, a Square with broken diagonal: $AB=x$? I am reading a geometry lecture note and it was given this figure below.
 Where ABCD is square of size $AB=x$. $BE =12$, $EF=3$ and $FD=9$.
Please help me to find the value of $x$.
I do not know how to start. So far I know that $BD =x\sqrt 2.$ 

 A: Strategies in the other answers apply to variants of the problem with arbitrary segment lengths (with some constraints). In this particular case, the relation $9+3=12$ invites a specialized argument that, as a bonus, helps explain why the solution is so nice. Consider this generalization:

Let's "complete the $q$-square": 

Observe that the $q$-square and the bounding $s$-square necessarily have the same center ... that is, the same center of symmetry. Thus, we can rotate all elements by $90^\circ$ about that point.

Finally, invoking Pythagoras,

$$p^2 + (p+q)^2 = s^2$$

In the problem at hand, we have
$$9^2 + 12^2 = 15^2$$
(The reader may recognize the key figure as a scaled version of our friend, the $3$-$4$-$5$ right triangle.)
A: $BD^2=(9+12)^2+3^2$, you just need to shift FD to the extension of BE. 
A: You can use Pythagorean theorem and find the square side.

A: Using the sine rule, cosine rule and pythagorean theorem will come in very handy here, consider the following facts:


*

*the distance $BF=\sqrt{12^2+3^2}$ lets call this distance $\alpha$

*given $\alpha$, the angle $\displaystyle \angle BFE=\sin^{-1}\left(\frac{12}{3\sqrt{17}}\right)$, call the angle $\beta$

*given $\beta$, the length $BD$ can be calculated by $BD^2=\alpha^2+9^2-2(\alpha)(9)\cos(\beta)$, call the length $\gamma$

*and lastly, given the length, the value of $x$ is given by $x=\sin(45)\cdot\gamma$


now, following these procedures, can you find numerical values for $\alpha$, $\beta$ and $\gamma$?

$$Edit$$
The sine rule states $$\frac{a}{\sin A}=\frac{b}{\sin B} = \frac{c}{\sin C}$$
Conversely, the cosine rule states $$a^2=b^2+c^2-2bc\cos A$$
And lastly, as you're probably well aware, the pythagorean theorem states
$$a^2=b^2+c^2$$
A: The picture is self explaining. The sum of the little square gives the big square.

A: The hint:
Use the following.
$$\measuredangle DFB=90^{\circ}+\arctan4,$$
$$DF=\sqrt{153},$$
$$FB=9$$ and
$$DB=x\sqrt2.$$
I got $x=15$.
