Problem 9.5. On the base $AC$ of isosceles triangle $ABC (AB = BC)$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.

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Answer: 17.

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Fig. 3: to the solution of problem 9.5

Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angle at vertex \(B\) is \(30^\circ\), so \(HM = \frac{1}{2} BM = 1.5\). Then \(AC = 2AH = 2(AM + MH) = 2 \cdot (7 + 1.5) = 17\).

![](https://cdn.mathpix.com/cropped/2024_05_06_ddcc40423d5e6a42b0ccg-33.jpg?height=234&width=713&top_left_y=963&top_left_x=370)

Fig. 4: to the solution of problem 9.5

Another solution. Mark a point \(K\) on \(MC\) such that \(\angle BKM = 60^\circ\) (Fig. 4; such a point lies exactly on the segment \(MC\) because \(\angle BCM = \angle BAM = \angle BMC - \angle ABM < 60^\circ\)). Note that in triangle \(BMK\), two angles are \(60^\circ\), so it is equilateral and \(BK = MK = BM = 3\). Also note that triangles \(ABM\) and \(CBK\) are equal because \(BC = AB\), \(\angle A = \angle C\), \(\angle AMB = \angle CKB = 120^\circ\) (therefore, \(\angle ABM = \angle CBK\)). Then \(CK = AM = 7\) and \(AC = AM + MK + KC = 7 + 3 + 7 = 17\).