# 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 $BD =x\sqrt 2.$

You can use Pythagorean theorem and find the square side.

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.)

• This is clever and well explain – Guy Fsone Oct 25 '17 at 6:27

$BD^2=(9+12)^2+3^2$, you just need to shift FD to the extension of BE.

• That's right! It's enough to add a new point G so that EFDG is a rectangle, and recall Pythagorean theorem to see the above answer. – CiaPan Oct 24 '17 at 15:41

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 $$a^2=b^2+c^2$$

The picture is self explaining. The sum of the little square gives the big square.

The hint:

Use the following.

$$\measuredangle DFB=90^{\circ}+\arctan4,$$ $$DF=\sqrt{153},$$ $$FB=9$$ and $$DB=x\sqrt2.$$

I got $x=15$.