In the diagram, $CE=CF=EF,\ EA=BF=2AB$, and $PA=QB=PC=QC=PD=QD=1$, Determine $BD$.
I tried working this question as : $$\angle ACB = \angle DQB = 12^\circ$$ So, by the cosine rule for $\triangle QBD$, $$q^2 = b^2 + d^2 - 2bd\cos Q = 1^2 + 1^2 - 2\cdot 1\cdot 1 \cdot \cos 12^\circ = 2-2\cos 12^\circ = 0.043704$$ Is this correct? Please advise.
Let $EF=5a$ and $CD\cap EF=\{K\}.$
Thus, $DK\perp AB$, $AB=a$ and by the Pythagoras theorem we obtain: $$AC=\sqrt{AK^2+CK^2}=\sqrt{\frac{AB^2}{4}+\frac{3EF^2}{4}}=a\sqrt{19}.$$ Now, $$\measuredangle ADC=\frac{1}{2}\left(2\measuredangle PAD+2\measuredangle PCD\right)=$$ $$=\frac{1}{2}\left(180^{\circ}-\measuredangle APD+180^{\circ}-\measuredangle CPD\right)=180^{\circ}-\frac{1}{2}\measuredangle APC.$$ which says $$\frac{1}{2}\measuredangle APC=180^{\circ}-\measuredangle ADC=\measuredangle ADK=\frac{1}{2}\measuredangle ADB,$$ which says that $\Delta ADB\sim\Delta APC,$ which gives $$\frac{BD}{AP}=\frac{AB}{AC}$$ or $$\frac{BD}{1}=\frac{a}{a\sqrt{19}}$$ or $$BD=\frac{1}{\sqrt{19}}.$$