prove that an isosceles triangle has the largest area

Question

Inside this angle, alpha is place in a segment of length whose ends are on the sides of the angle so that the area of ABC was maximum

Answer 1

Let the sides meeting at an angle $A$ have lengths $x,\,y$. You want to prove that, subject to $x^2+y^2-2xy\cos A=a^2$, the maximum of $\frac{1}{2}xy\sin A$ (or equivalently $xy$) takes $x=y$. So use the Lagrangian $L=xy+\lambda (x^2+y^2-2xy\cos A-a^2)$. Then $0=\partial_x L=y+\lambda (2x-2y\cos A)$, i.e. $2\lambda x+(1-2\lambda\cos A)y=0$. Similarly, $2\lambda y+(1-2\lambda\cos A)x=0$. Subtracting, $(x-y)(2\lambda (1+\cos A)-1)=0$. If $x\ne y$, $1-2\lambda\cos A=2\lambda$ so $2\lambda (x+y)=0$, whence $y=-x$. But we require $x,\,y>0$ to maximise area.