[m]c^2_{1}=a^2+l^2-2alcos φ [/m]; [m] 2φ = ∠ C[/m]
[m]c^2_{2}=b^2+l^2-2blcos φ [/m];
⇒ [m]\frac{a^2+l^2-c^2_{1}}{2al}=\frac{b^2+l^2-c^2_{2}}{2bl}[/m] ⇒
[m]\frac{a^2+l^2-c^2_{1}}{a}=\frac{b^2+l^2-c^2_{2}}{b}[/m] ⇒[m] b(a^2+l^2-c^2_{1})=a(b^2+l^2-c^2_{2})[/m] ⇒
[m]bl^2-al^2=ab^2-a^2b-bc^2_{1}+ac^2_{2}[/m]
[m]l^2\cdot (b-a)=ab(b-a)-bc^2_{1}+ac^2_{2}[/m]
По свойству биссектрисы:
[m]\frac{c_{1}}{c_{2}}=\frac{a}{b}[/m] ⇒ [m]bc_{1}=ac_{2}[/m]
[m]l^2\cdot (b-a)=ab(b-a)-bc_{1}\cdot c_{1}+ac_{2}\cdot c_{2}[/m]
[m]l^2\cdot (b-a)=ab(b-a)-ac_{2}\cdot c_{1}+bc_{1}\cdot c_{2}[/m]
[m]l^2\cdot (b-a)=ab(b-a)-c_{1}c_{2}\cdot (b-a)[/m]
[m]l^2=ab-c_{1}c_{2}[/m]