If the bisector of the angle of a triangle $ABC$ meet the opposite sides in $A^{\text{'}}, B^{\text{'}}$ and $C^{\text{'}}$, prove that the ratio of the areas of the triangles $A^{\text{'}}{B}^{\text{'}}{C}^{\text{'}} \& ABC$ is $2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}:\cos \frac{A-B}{2}\cos \frac{B-C}{2}\cos \frac{C-A}{2}$.

In triangle $ABC$ in the sides $BC, CA, AB$ are taken three points $A^{\text{'}}, B^{\text{'}}, C^{\text{'}}$ such that $B{A}^{\text{'}}:A^{\text{'}}C=C{B}^{\text{'}}:B^{\text{'}}A=A{C}^{\text{'}}:C^{\text{'}}B=m:n$;

Prove that if $A{A}^{\text{'}}, B{B}^{\text{'}}$ and $C{C}^{\text{'}}$ be joined they will form by their intersections a triangle whose area is to that of the triangle $ABC$ as ${\left(m-n\right)}^2:m^2+mn+n^2$.

The circle inscribed in the triangle $ABC$ touches the sides $BC, CA$ and $AB$ in the points $A_1, B_1$ and ${\mathrm{C}}_1$, respectively. Similarly, the circle inscribed in the triangle $A_1{B}_1{C}_1$ touches the sides in $A_2, B_2, C_2$, respectively and so on; if $A_n{B}_n{C}_n$ be the $n^{th}$ triangle so formed, prove that its angles are $\frac{\mathrm{\pi }}{3}+{\left(-2\right)}^{-n}\left(A-\frac{\mathrm{\pi }}{3}\right), \frac{\mathrm{\pi }}{3}+{\left(-2\right)}^{-n}\left(B-\frac{\mathrm{\pi }}{3}\right)$ and $\frac{\mathrm{\pi }}{3}+{\left(-2\right)}^{-n}\left(C-\frac{\mathrm{\pi }}{3}\right)$.

Hence prove that the triangle so formed is ultimately equilateral.