Пусть AB = BC = b, AC = a, AK = L, ∠A = ∠C = 2X. Тогда:
\frac{AB}{AC} = \frac{BC - KC}{KC} => \frac{b}{a} = \frac{b-KC}{KC} => b*KC = ab - a*KC => KC = \frac{ab}{a+b}\\ \Delta AKC: AK^2 = AC^2 + KC^2 - 2*AC*KC*cosC => l^2 = a^2 + (\frac{ab}{a+b})^2 - 2*a*\frac{ab}{a+b}*\frac{a}{2b} => l^2 = a^2 + (\frac{ab}{a+b})^2 - \frac{a^3}{a+b} => l^2 = \frac{a^4 + 2a^3b + a^2b^2 + a^2b^2 - a^4-a^3b}{(a+b)^2} =>" alt="\Delta ABH: cosA = \frac{AH}{AB} = \frac{a}{2b} = cosC\\ \Delta ABC: \frac{AB}{AC} = \frac{BK}{KC} => \frac{AB}{AC} = \frac{BC - KC}{KC} => \frac{b}{a} = \frac{b-KC}{KC} => b*KC = ab - a*KC => KC = \frac{ab}{a+b}\\ \Delta AKC: AK^2 = AC^2 + KC^2 - 2*AC*KC*cosC => l^2 = a^2 + (\frac{ab}{a+b})^2 - 2*a*\frac{ab}{a+b}*\frac{a}{2b} => l^2 = a^2 + (\frac{ab}{a+b})^2 - \frac{a^3}{a+b} => l^2 = \frac{a^4 + 2a^3b + a^2b^2 + a^2b^2 - a^4-a^3b}{(a+b)^2} =>" align="absmiddle" class="latex-formula">
l = \frac{a}{a+b}\sqrt{2b^2+ab}" alt="l^2 = \frac{a^3b+2a^2b^2}{(a+b)^2} = \frac{a^2b(a+2b)}{(a+b)^2} => l = \frac{a}{a+b}\sqrt{2b^2+ab}" align="absmiddle" class="latex-formula">