1. **Problem statement:** Given a circle $(O; R)$ with diameter $AB$, tangents $(d)$ at $A$ and $(d')$ at $B$. A line through $O$ intersects $(d)$ at $M$ and $(d')$ at $P$. From $O$, draw a perpendicular to $MP$ intersecting $(d')$ at $N$. Let $OI \perp LMN$ at $I$. Prove: (a) $OM = OP$ and $\triangle NMP$ is isosceles. 2. **Formula and rules:** - Tangent lines to a circle are perpendicular to the radius at the point of tangency. - The power of a point and properties of tangents and chords in circles. 3. **Proof of (a):** - Since $(d)$ and $(d')$ are tangents at $A$ and $B$, $OA \perp (d)$ and $OB \perp (d')$. - $AB$ is diameter, so $O$ is midpoint of $AB$. - Line through $O$ intersects $(d)$ at $M$ and $(d')$ at $P$. - Triangles $OMA$ and $OPB$ are right triangles with right angles at $A$ and $B$ respectively. - Because $OA = OB = R$ and $OM$, $OP$ are segments from $O$ to tangents, by symmetry and equal tangents from $O$, $OM = OP$. - In $\triangle NMP$, since $ON \perp MP$ and $N$ lies on $(d')$, $\angle NMP = \angle NPM$, so $\triangle NMP$ is isosceles. Final answer for (a): $$OM = OP \quad \text{and} \quad \triangle NMP \text{ is isosceles}.$$