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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 $BD =x\sqrt 2.$

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You can use Pythagorean theorem and find the square side. enter image description here

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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:

enter image description here

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

enter image description here

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.

enter image description here

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.)

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    $\begingroup$ This is clever and well explain $\endgroup$ – Guy Fsone Oct 25 '17 at 6:27
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$BD^2=(9+12)^2+3^2$, you just need to shift FD to the extension of BE.

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  • $\begingroup$ 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. $\endgroup$ – CiaPan Oct 24 '17 at 15:41
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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 $$a^2=b^2+c^2$$

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

enter image description here

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The hint:

Use the following.

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

I got $x=15$.

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