Atrisinājums

Apzīmējam \(E C=x, \sphericalangle B A D=\sphericalangle D A E=\alpha, \sphericalangle E A C=2 \alpha\) un atliekam punktiem \(D\) un \(E\) attiecīgi simetriskus punktus \(D_{1}\) un \(E_{1}\) pret taisni \(A B\) (skat. 9. att.). Tātad \(B D_{1}=B D=3\) un \(D E=D_{1} E_{1}=4\). Izmantosim bisektrises īpašību un nogriežṇu garumu vienādību \(A E_{1}=A E\) (simetrijas dēļ):
(1) \(\triangle B A C\) bisektrise \(A E: \quad \frac{A B}{A C}=\frac{B E}{E C}=\frac{7}{x}\); (2) \(\triangle B A E\) bisektrise \(A D: \quad \frac{A B}{A E}=\frac{B D}{D E}=\frac{3}{4}\); (3) \(\triangle E_{1} A C\) bisektrise \(A D: \frac{A E_{1}}{A C}=\frac{E_{1} D}{D C} \Rightarrow \frac{A E}{A C}=\frac{10}{4+x}\).

No (2) un (3) iegūstam, ka

\[\frac{A B}{A C}=\frac{A B}{A E} \cdot \frac{A E}{A C}=\frac{3}{4} \cdot \frac{10}{4+x}=\frac{30}{4(4+x)} .\]

\[\frac{30}{4(4+x)}=\frac{7}{x} \quad \Rightarrow \quad 30 x=112+28 x \quad \Rightarrow \quad x=56\]