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

euclidean-geometrytriangles

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

enter image description here

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 $x$.

Best Answer

enter image description here 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.

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