parmenides51 wrote: In triangle $ABC$, the bisector $CL$ was drawn. The incircles of triangles $CAL$ and $CBL$ touch $AB$ at points $M$ and $N$ respectively. Points $M$ and $N$ are marked on the picture, and then the whole picture except the points $A, L, M$, and $N$ is erased. Restore the triangle using a compass and a ruler. (V.Protasov) Let $I, I_a, I_b$ be the incentresof $\triangle ABC, \triangle ACL, \triangle BCL$ respectively. Let $D$ be the foot of perpendicular from $I$ to $AB$. Claim : $MD = LN$ Proof : Note that $$ MD = \frac{(AC + BC - AC) - (AL + AC - LC)}{2} = \frac{BL + LC - BC}{2} = LN \quad \square$$ Now first construct $D$ by using our claim. Then construct $I, I_a, I_b$ in that order. Construct the incircles of $\triangle ACL, \triangle BCL$. Now construct $C = LI \cap \text{A-tangent to incircle of }\triangle ACL$. Then construct $B = II_b \cap AL$. Now $\triangle ABC$ is restored. $\blacksquare$