7.1. In an acute-angled triangle $A B C$, the median $B M$ and the altitude $C H$ were drawn. It turned out that $B M=C H=\sqrt{3}$, and $\angle M B C=\angle A C H$. Find the perimeter of triangle $A B C$.
Answer: 6.

Solution. We will prove that triangle $ABC$ is equilateral, with all sides equal to 2. Drop a perpendicular $MK$ from point $M$ to side $AB$. In triangle $BMK$, the leg $KM$ is half the hypotenuse $MB$, so $\angle KBM = 30^{\circ}$. Let $O$ be the intersection of $MB$ and $CH$, then $\angle MOC = 60^{\circ}$. Denote $\angle MBC = \angle ACH = \alpha$, then $\angle HCB = 60^{\circ} - \alpha$, so $\angle BCA = 60^{\circ}$. Therefore, triangles $MOC$ and $MCB$ are similar by two angles. From the similarity, we get $MC^2 = BM \cdot OM = CH \cdot OM$. By the Law of Sines, $OM = MC \cdot \sin \alpha / \sin 60^{\circ}$. From triangle $AHC$, we find $HC = 2MC \cdot \cos \alpha$. Substituting this into the formula $MC^2 = CH \cdot OM$ and canceling $MC^2$, we get $\sin 2\alpha = \sqrt{3} / 2$, from which $\alpha = 30^{\circ}$. Therefore, triangle $ABC$ is equilateral.