Triangle $ABC$ has $AB=BC=10$ and $CA=16$. The circle $\Omega$ is drawn with diameter $BC$. $\Omega$ meets $AC$ at points $C$ and $D$. Find the area of triangle $ABD$.
1. **Identify the given information and draw the necessary elements:**
   - Triangle \(ABC\) has sides \(AB = BC = 10\) and \(CA = 16\).
   - Circle \(\Omega\) is drawn with diameter \(BC\).
   - \(\Omega\) meets \(AC\) at points \(C\) and \(D\).

2. **Label the midpoint of \(BC\) as \(M\):**
   - Since \(BC\) is the diameter of the circle, \(M\) is the center of the circle.
   - Therefore, \(M\) is the midpoint of \(BC\), and \(BM = MC = 5\).

3. **Use the properties of the circle and similar triangles:**
   - Since \(BC\) is the diameter, by Thales' Theorem, \(\angle BDC = 90^\circ\).
   - This implies that \(\triangle BDC\) is a right triangle with \(\angle BDC = 90^\circ\).

4. **Determine the length of \(CD\):**
   - Since \(M\) is the midpoint of \(BC\), \(MD = MC = 5\).
   - \(\triangle MDC\) is a right triangle with \(MD = 5\) and \(MC = 5\).
   - By the Pythagorean theorem, \(CD = \sqrt{MC^2 - MD^2} = \sqrt{5^2 - 5^2} = \sqrt{0} = 0\).
   - This is incorrect; instead, we should use the fact that \(\triangle MDC \sim \triangle BAC\).

5. **Correct the similarity approach:**
   - Since \(\triangle MDC \sim \triangle BAC\), the ratio of corresponding sides is the same.
   - Let \(CD = x\). Then, \(\frac{CD}{BC} = \frac{AC}{AB}\).
   - \(\frac{x}{10} = \frac{16}{10}\).
   - Solving for \(x\), we get \(x = 16\).

6. **Calculate the area of \(\triangle ABD\):**
   - Since \(\angle BDC = 90^\circ\), \(\angle BDA = 90^\circ\).
   - \(\triangle BDA\) is a right triangle with \(\angle BDA = 90^\circ\).
   - We need to find \(BD\) and \(AD\).
   - Using the Pythagorean theorem in \(\triangle BDC\), \(BD = \sqrt{BC^2 - CD^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6\).
   - The area of \(\triangle ABD\) is \(\frac{1}{2} \cdot BD \cdot AD = \frac{1}{2} \cdot 6 \cdot 8 = 24\).

The final answer is \(\boxed{24}\).