Let ABC be a triangle, which biggest (in other case is same) angle is <ACB. Let CH is a perpendicular to AB, H lie on the side AB. Let <ACH=ф. In triangle AHC we have tga=cotgф =>atgatgф=a and in triangle BCH btgbtg(y-ф)=b => a+b=atgatgф+btgbtg(y-ф). In this broblem we have a+b=tg(y/2)(atga+btgb) => tg(y/2)(atga+btgb)=atgatgф+btgbtg(y-ф). atga(tgф-tg(y/2))=btgb(tg(y/2)-tg(y-ф)). We use that tgx-tgy=(sin(x-y))/cosxcosy => atgasin(ф-y/2)/cosфcos(y/2)=btgbsin(ф-y/2)/cos(y-ф)cos(y/2) => atga/cosф=btgb/cos(y-ф). But in triangle AHC cosф=sina and cos(y-ф)=sinb. We use sinus law: a/sina=b/sinb and we have tga=tgb. Is it OK?

$\left(A+\frac C2\right)+\left(B+\frac C2\right)=180^{\circ}\Longrightarrow \cos \left(A+\frac C2\right)=-\cos \left(B+\frac C2\right).$ $a+b=\tan \frac C2\cdot (a\tan A+b\tan B)\Longleftrightarrow a\left( \cot \frac C2-\tan A\right)=b\left(\tan B-\cot \frac C2\right)\Longleftrightarrow$ $\frac{a\cos \left(A+\frac C2\right)}{\sin \frac C2\cos A}=\frac{b\cos \left(B+\frac C2\right)}{\sin \frac C2\cos B}\Longleftrightarrow$$\cos \left(A+\frac C2\right)=0\ \ \vee\ a\cos B=b\cos A\Longleftrightarrow$ $A+\frac C2=B+\frac C2\ \vee\ \tan A=\tan B\Longleftrightarrow A=B.$

DPopov wrote: Let $ a,b,c$ be the lengths of the sides of a triangle, and $ \alpha, \beta, \gamma$ respectively, the angles opposite these sides. Prove that if \[ a + b = \tan{\frac {\gamma}{2}}(a\tan{\alpha} + b\tan{\beta})\] the triangle is isosceles. The way I did it was this: Let $ x = \tan \frac {\alpha}{2}, y = \tan \frac {\beta}{2}$. Then: $ a = 2R \cdot \sin a = 2R \frac {2x}{1 + x^2}, b = 2R \frac {2y}{1 + y^2}$ $ \tan \alpha = \frac {2x}{1 - x^2}, \tan \beta = \frac {2y}{1 - y^2}$ $ \tan \frac {\gamma}{2} = \frac {1 - xy}{x + y}$ Then it is equivalent to: $ (x + y)^2(1 - x^2)(1 - y^2) = 2(1 - xy)^2(x^2 + y^2)$ And: $ (x + y)^2 \le 2(x^2 + y^2)$ with equality iff $ x = y$ $ (1 - x^2)(1 - y^2) \le (1 - xy)^2$ with equality iff $ x = y$. So: $ RHS \ge LHS$ with equality iff $ \tan \frac {\alpha}{2} = \tan \frac {\beta}{2} \iff \alpha = \beta \iff a = b$. Since there is equality we can conclude that the triangle is isosceles.

my solution = let angles of triangle be $A,B,C$ respective to side lengths $a,b,c$ wlog let $\angle A\ge \angle B$ and hence $a\ge b$ so by thchebytcheff we have $2(a+b)= 2\tan {\frac{C}{2}}(a\tan A +b\tan B)\ge \tan {\frac{C}{2}}(a+b)(\tan A+\tan B)$ so $2\ge \tan {\frac{C}{2}}(\tan A+\tan B)$ or $2cosAcosB\ge \tan {\frac{C}{2}}(sin(A+B))=\tan {\frac{C}{2}}(sinC)$ or we get, $cos(A-B)-cosC\ge 2(\sin\frac{C}{2})^2\leftrightarrow cos(A-B)\ge 1$ and thus $\angle A=\angle B$ so $ABC$ is isosceles triangle. we are done