1. **Problem statement:** We have two concentric circles with center $O$. Points $A, B, C, D$ lie on the outer circle, and $E$ lies on the inner circle. Lines $\overline{ABCD}$ and $\overline{AEO}$ are straight. $M$ is the midpoint of $AD$. Given: $OE=25$ cm, $AE=14$ cm, and $AB=16$ cm. We need to find the length $OM$. 2. **Understanding the problem:** Since $O$ is the center, $OE$ is a radius of the inner circle, so $OE=25$ cm is the inner radius. $AB=16$ cm is a chord length on the outer circle. $M$ is midpoint of $AD$, so $AM=MD$. 3. **Step 1: Find $AD$ using $AB$ and the fact that $ABCD$ is a straight line.** Since $ABCD$ is a straight line, $AD = AB + BC + CD$. But we don't have $BC$ or $CD$ lengths. However, $M$ is midpoint of $AD$, so $AM = MD = \frac{AD}{2}$. 4. **Step 2: Use the right triangle $AEO$ to find $OE$.** Given $AE=14$ cm and $OE=25$ cm, and $AEO$ is a straight line, so $AO = AE + EO = 14 + 25 = 39$ cm. 5. **Step 3: $AO$ is radius of outer circle, so outer radius $R=39$ cm. Inner radius $r=25$ cm.** 6. **Step 4: Use $AB=16$ cm chord length on outer circle to find distance from center $O$ to chord $AB$.** The perpendicular distance from center $O$ to chord $AB$ is $d$. Using chord length formula: $$AB = 2\sqrt{R^2 - d^2}$$ Substitute $AB=16$, $R=39$: $$16 = 2\sqrt{39^2 - d^2}$$ Divide both sides by 2: $$8 = \sqrt{1521 - d^2}$$ Square both sides: $$64 = 1521 - d^2$$ Rearranged: $$d^2 = 1521 - 64 = 1457$$ So, $$d = \sqrt{1457}$$ 7. **Step 5: $d$ is the perpendicular distance from $O$ to chord $AB$. Since $ABCD$ is a straight line, $M$ lies on $AD$ which is on the same line. We want $OM$, the distance from $O$ to midpoint $M$ of $AD$.** 8. **Step 6: Since $M$ is midpoint of $AD$, and $ABCD$ is a straight line, $OM$ is the distance from $O$ to $M$ along the line $ABCD$.** 9. **Step 7: Use the fact that $AB=16$ and $AO=39$ to find $AM$.** Since $AB=16$, and $AO=39$, and $ABCD$ is a straight line, $A$ and $B$ lie on the outer circle. The length $AD$ is twice $AM$. 10. **Step 8: Use the right triangle formed by $O$, $M$, and the perpendicular from $O$ to $ABCD$.** Since $d = \sqrt{1457}$ is the perpendicular distance from $O$ to chord $AB$, and $M$ lies on $ABCD$, the distance $OM$ along the line is: $$OM = \sqrt{R^2 - (\frac{AD}{2})^2}$$ But we need $AD$. 11. **Step 9: Use the fact that $M$ is midpoint of $AD$ and $AB=16$ to find $AD$.** Since $ABCD$ is a straight line, and $AB=16$, $BC$ and $CD$ are unknown, so we cannot find $AD$ directly. But $M$ is midpoint of $AD$, so $AM=MD$. 12. **Step 10: Use the fact that $M$ lies on the line $ABCD$ and $OM$ is the distance from $O$ to $M$. Since $OE=25$ and $AE=14$, and $AO=39$, $OM$ is the average of $OE$ and $AO$ because $M$ is midpoint of $AD$ which includes $A$ and $D$ on the outer circle.** 13. **Step 11: Calculate $OM$ as midpoint between $OE=25$ and $AO=39$:** $$OM = \frac{OE + AO}{2} = \frac{25 + 39}{2} = \frac{64}{2} = 32$$ **Final answer:** $$\boxed{OM = 32 \text{ cm}}$$