<s>[INST] Basic geometry: find the angle Question: Find $\angle BDE$ when
$\angle A=20^\circ$, $\overline{AB}=\overline{AC}$, $\angle CBD=65^\circ$, and $\angle BCE=25^\circ$.

So far I figured out that $\angle BDE=\angle AED-15^\circ$ and $\overline{BD}\bot\overline{CE}$ that don't seem to be helping.
 [/INST] This is a variation of the notorious Langley's problem. 
We follow a trigonometric approach. Let $\angle BDE=x$.
i) Use the Law of sines in $\triangle EBC$, 
$$\frac{\sin 25^\circ}{EB}=\frac{\sin75^\circ}{AB}.$$
ii) Use the Law of sines in $\triangle DBC$, 
$$\frac{\sin 80^\circ}{BD}=\frac{\sin 35^\circ}{AB}.$$
iii) Use the Law of sines in $\triangle EBD$, 
$$\frac{\sin x}{EB}=\frac{\sin (165^\circ-x)}{BD}=\frac{\sin (15^\circ+x)}{BD}.$$
It follows that
$$\frac{\sin (15^\circ+x)}{\sin x}=\frac{\sin 80^\circ}{\sin 35^\circ}\cdot  \frac{\sin 75^\circ}{\sin 25^\circ}$$
Since $\sin (15^\circ+x)=\sin 15^\circ\cos x+\cos 15^\circ\sin x$, 
we obtain that
$$\cot x=\frac{1}{\sin 15^\circ}\cdot\frac{\sin 80^\circ}{\sin 35^\circ}\cdot  \frac{\sin 75^\circ}{\sin 25^\circ}-\cot 15^\circ.$$
Therefore $x=5^\circ$.
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