A geometry problem by safa m

Let $NK, MB$ and $MC$ be tangent to the circle at $A, B$ and $C$ respectively. Now, we know that $AN = BN, AK = CK$ and $MB = MC.$

Therefore, $MB = 5.$ So our answer is $BC = s - MB = 9 - 5 = \boxed {4}$

we put a dot on every circle tangent. the left one E , the up one F , the right one D , and the down one A. we know that FK=KD , NF=NE , BA=BE , DC=CA , and we know MNK=10 , it means that MN+NF+MK+KF=10 and ===> MN+EN+MK+KD=10 and we know that MBC=18 so we have ====> EB+BA+AC+DC=18-10=8 Thus===>2BA+2AC=8 ===> BA+AC=4 ===> BC=4