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**SlipEternal** If $m(D)>2m(B)$ then consider if $m(D) = 2m(B) = 2\theta$. This gives the angles $m(A) = m(D) = 2\theta$ and $m(B)=m(C) = \theta$, so $\theta+\theta+2\theta+2\theta = 6\theta = 2\pi=360^\circ$. Thus, $\theta = \dfrac{\pi}{3} = 60^\circ$. Now, you know that $2m(B) < 2\theta$ so $m(B) < \theta$. You should know the sides of a 30-60-90 triangle are $a, a\sqrt{3}, 2a$. So, since the hypotenuse is 5, we know $|CK|>\dfrac{5}{2}$. So, $|BC| > \dfrac{5}{2} + 3 + \dfrac{5}{2} = 8$. Therefore, the smallest integer value for $|BC|$ would be 9.