Let $ABCD$ be a quadrilateral with $\overline{AB}\parallel\overline{CD}$, $AB=16$, $CD=12$, and $BC<AD$. A circle with diameter $12$ is inside of $ABCD$ and tangent to all four sides. Find $BC$.
1. Let the circle have center \( O \) and \( W, X, Y, Z \) be the points of tangency of the circle with \( AB, BC, CD, DA \) respectively. Notice that \(\angle OWB = \angle OYC = 90^\circ\), so \( WOY \) is a straight line and the altitude of the trapezoid. 

2. Since the circle is tangent to all four sides of the quadrilateral, the sum of the lengths of opposite sides of the quadrilateral is equal. Therefore, we have:
   \[
   AB + CD = AD + BC
   \]
   Given \( AB = 16 \) and \( CD = 12 \), we get:
   \[
   16 + 12 = AD + BC \implies 28 = AD + BC
   \]

3. Let \( M \) be the foot of the altitude from \( D \) to \( AB \) and \( N \) be the foot of the altitude from \( C \) to \( AB \). Translate \(\triangle BCN\) to \(\triangle B'C'N'\) such that \( C'N' \) coincides with \( DM \). We know \( AB' = 16 - 12 = 4 \), and \( AD + DB' = AD + BC = 28 \).

4. Using Heron's formula for the area of \(\triangle AB'D\), we have:
   \[
   \text{Area} = \sqrt{s(s-AB')(s-AD)(s-DB')}
   \]
   where \( s \) is the semi-perimeter:
   \[
   s = \frac{AB' + AD + DB'}{2} = \frac{4 + 28}{2} = 16
   \]
   Let \( AD = x \), so \( DB' = 28 - x \). The area of \(\triangle AB'D\) is also given by:
   \[
   \text{Area} = \frac{1}{2} \times AB' \times CD = \frac{1}{2} \times 4 \times 12 = 24
   \]

5. Equating the two expressions for the area, we get:
   \[
   24 = \sqrt{16(16 - 4)(16 - x)(16 - (28 - x))}
   \]
   Simplifying inside the square root:
   \[
   24 = \sqrt{16 \cdot 12 \cdot (16 - x) \cdot (x - 12)}
   \]
   Squaring both sides:
   \[
   576 = 192(16 - x)(x - 12)
   \]
   Dividing both sides by 192:
   \[
   3 = (16 - x)(x - 12)
   \]

6. Solving the quadratic equation:
   \[
   3 = 16x - x^2 - 192 + 12x \implies x^2 - 28x + 195 = 0
   \]
   Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{28 \pm \sqrt{784 - 780}}{2} = \frac{28 \pm 2}{2}
   \]
   \[
   x = 15 \quad \text{or} \quad x = 13
   \]

7. Since \( BC < AD \), we have \( BC = 13 \) and \( AD = 15 \).

The final answer is \( \boxed{13} \).