Atrisinājums

Apzīmējam \(\sphericalangle B C D=2 \alpha\) un \(\sphericalangle A B C=2 \beta\). No tā, ka četrstūrim \(A B C D\) var apvilkt riṇḳa līniju, iegūstam, ka \(\sphericalangle BAD=180^{\circ}-2 \alpha\) un \(\sphericalangle ADC=180^{\circ}-2 \beta\). Uz malas \(AD\) atliek punktu \(E\) tā, ka \(A B=A E=a\) un \(D E=D C=b\) (skat. 13. att.). Tad \(\triangle ABE\) un \(\triangle CDE\) ir vienādsānu trijstūri, kuriem \(\sphericalangle AEB=\sphericalangle ABE=\alpha\) un \(\sphericalangle CED = \sphericalangle ECD=\beta\). Izmantojot kosinusu teorēmu trijstūrī \(ABE\), iegūstam, ka

\[\begin{gathered} BE^{2} = AB^{2} + AE^{2} - 2AB \cdot AE \cdot \cos \left(180^{\circ} - 2 \alpha\right) ; \\ BE^{2} = a^{2}+a^{2}+2a^{2} \cos 2 \alpha=2 a^{2}(1+\cos 2 \alpha)=2 a^{2} \cdot 2 \cos ^{2} \alpha ; \\ BE = 2a \cdot \cos \alpha . \end{gathered}\]

Pēc sinusu teorēmas trijstūrī \(B C E\) iegūstam, ka

\[ \begin{equation*} \frac{B C}{\sin \left(180^{\circ}-\alpha-\beta\right)}=\frac{E B}{\sin (2 \alpha-\beta)}=\frac{2 a \cdot \cos \alpha}{\sin (2 \alpha-\beta)} . \tag{1} \end{equation*} \]

 Tagad novelkam leṇku \(ABC\) un \(BCD\) bisektrises, kas krustojas kādā punktā \(F\) (skat. 14. att.). Bisektrises \(BF\) krustpunktu ar \(AC\) apzīmējam ar \(G\). Tad \(\sphericalangle AGB = 180^{\circ}-\sphericalangle BAG-\sphericalangle ABG = 180^{\circ}-\left(180^{\circ}-2 \alpha\right)-\beta=2 \alpha-\beta\). No sinusu teorēmas trijstūrī \(ABG\) iegūstam, ka

\[\begin{equation*} \frac{a}{\sin (2 \alpha-\beta)} = \frac{BG}{\sin \left(180^{\circ} - 2 \alpha\right)} = \frac{B G}{\sin 2 \alpha} . \tag{2} \end{equation*}\]

No leņḳu sakarībām trijstūrī \(BCF\) leņḳu sakarībām izriet, ka \(\sphericalangle BFC = 180^{\circ} - \alpha - \beta = \sphericalangle BEC\) un

\[\frac{BC}{\sin \left(180^{\circ}-\alpha-\beta\right)} = \frac{B F}{\sin \alpha} \quad \stackrel{(1)}{\Rightarrow} \quad \frac{2 a \cdot \cos \alpha}{\sin (2 \alpha-\beta)} = \frac{B F}{\sin \alpha} \quad \Rightarrow\]

\[\Rightarrow \quad \frac{a}{\sin (2 \alpha-\beta)} = \frac{BF}{\sin 2 \alpha} \quad \stackrel{(2)}{\Rightarrow} \quad \frac{BF}{\sin 2 \alpha}=\frac{BG}{\sin 2 \alpha} \quad \Rightarrow \quad BF=BG.\]

Tas nozīmē, ka punkti \(F\) un \(G\) sakrīt un atrodas uz \(AD\).