Finding the angle $x$ in $\triangle ABC$

Question

What is the angle $x$ in the following diagram.

I could solve this problem by applying the Sine Rule for $\triangle ABD$ and $\triangle BDC$. After some simplification I got $$\sin^2 x =\sin(45-x)\sin(135-x) \Rightarrow \sin^2x = \cos(x-45)\cos(x+45)$$ And it was not hard to solve it and I got $x = 30^\circ$.

I'm wondering if it is possible to solve this problem with other approaches (especially by using Euclidean Geometry techniques). For this purpose I drew a parallel line to $AB$ from the point $D$ and since $D$ is mid-point of $AC$, the intersection to $BC$ also is the midpoint of $BC$. But unfortunately I couldn't find the angle $x$.

Answer 1

We extend $BD$ to $B'$ to make that $BD=DB'$.Then we connect $B'C$ and draw the perpendicular line of $AC$ through point $B'$,which insects with E.

It's easy to find that $\triangle{ABD}\approxeq\triangle{DB'C}$ and $\triangle{DB'C}\thicksim\triangle{BB'C}$

To simplify the statement,we let $EB'=1$,then$DB'=BD=\sqrt{2}$(isosceles right triangle). And we find that$B'C=2$,based on the similar triangle theory.Then we can get $x=30^\circ$.

Answer 2

There are some observations you can make on the triangle that will render the application of the sine rule trivial, i.e. you can do most of the problem in "geometry land" and leave only a tiny bit for the "algebra land".

Note that after drawing a segment from $D$ parallel to $AB$, you get two triangles $ABD,BDE$ with the same height but base segments $AB:DE=2:1$, hence their areas also have a ratio $2:1$:

But they are similar ($ADB\cong BED$) because $\angle DAB=x=\angle EBD$ and $\angle ADB=45=\angle BED$ (the latter reqiures some angle chasing). The areas are of ratio $2:1$, hence respective sides have a ratio of $\sqrt 2 : 1$. That is, $AD = \sqrt 2 EB$ and so $AC=\sqrt 2 BC$.

I don't see how I can proceed further without the sine rule, but now it's trivial (writing single letters only for the angles of $\triangle ABC$): $$\frac{\sin A}{BC}\overset{\textrm{sine rule}}{=}\frac{\sin B}{AC}\overset{\textrm{by above}}{=}\frac{\sin 135}{\sqrt 2 BC}=\frac 1 {2 BC}\Rightarrow\sin A=\frac 1 2\Rightarrow x=30.$$

You may find a way to geometrize the last step, too, but I couuldn't see how in a reasonably short amount of time :)

Answer 3

As in the figure above, draw from $C$ the line parallel to $BD$, and let $E$ be its intersection with $AB$. Drop from $E$ the perpendicular to $BC$, intersecting $BC$ in $H$.

Let for simplicity $\overline{BD} = 1$. Since $AB \cong BE$ and $\overline{EC}= 2$ (Thales), by the similarity $\triangle ABD \sim \triangle CBE$, we get $\overline{BE} = \sqrt 2$. Note that $\measuredangle EBC = 45^\circ$. Hence $\overline{EH} = 1 = \frac12 \overline{EC}$. From which we derive our conclusion.