Subject 3. On the plane of isosceles triangle $\mathrm{ABC}$, perpendicular $\mathrm{AM}$ is raised. If $\mathrm{AM}=\mathrm{AC}=\mathrm{b}, \mathrm{AB}=\mathrm{c}$ and $\left(b^{2}+c^{2}-6\right)^{2}+\left(b^{2}-c^{2}+4\right)^{2}=0$, calculate the distance from M to BC.
## Subject 3.

|  | $\left(b^{2}+c^{2}-6\right)^{2}+\left(b^{2}-c^{2}+4\right)^{2}=0 \Rightarrow c=\sqrt{5}$ and $b=1$ |  |
| :--- | :--- | :---: |
| $\mathrm{AC}=1, \mathrm{AB}=\sqrt{5} \Rightarrow \mathrm{BC}=\sqrt{5}(1+1<\sqrt{5})$ | $\mathbf{2 p}$ |  |
|  | Construction of the perpendicular MN to $\mathrm{BC}(\mathrm{T} 3 \perp)$ | $\mathbf{2 p}$ |
|  | Calculation of the height $A N=\frac{\sqrt{95}}{10}$ | $3 \mathbf{p}$ |
|  | Calculation of the distance $M N=\frac{\sqrt{195}}{10}$ |  |