1. **Problem statement:** We have two secant segments $\overline{IT}$ and $\overline{MT}$ intersecting outside a circle at point $T$. Points $H$ and $U$ lie on the circle and on segments $\overline{IT}$ and $\overline{MT}$ respectively. Given $IT = 21$ cm, $IH = 8$ cm, and $\overline{MU} \cong \overline{UT}$, find $MT$. 2. **Relevant theorem:** For two secants intersecting outside a circle, the product of the whole secant segment and its external segment is equal for both secants: $$IT \times IH = MT \times MU$$ where $IH$ and $MU$ are the external segments from point $T$ to the circle. 3. **Given data:** - $IT = 21$ cm (whole secant segment) - $IH = 8$ cm (external segment of $IT$) - $\overline{MU} \cong \overline{UT}$ means $MU = UT$ 4. **Find:** $MT$, the length of the secant segment $\overline{MT}$. 5. **Analyze $MT$:** Since $MU = UT$, point $U$ is the midpoint of $\overline{MT}$. Therefore, $$MU = UT = \frac{MT}{2}$$ 6. **Apply the secant segment theorem:** $$IT \times IH = MT \times MU$$ Substitute known values and expressions: $$21 \times 8 = MT \times \frac{MT}{2}$$ Simplify: $$168 = \frac{MT^2}{2}$$ Multiply both sides by 2: $$336 = MT^2$$ 7. **Solve for $MT$:** $$MT = \sqrt{336} = \sqrt{16 \times 21} = 4\sqrt{21} \approx 18.33 \text{ cm}$$ **Final answer:** $$\boxed{MT = 4\sqrt{21} \approx 18.33 \text{ cm}}$$