15 Quadrilateral ABCD is inscribed in circle $$ Q . A B=12 . B C=16 $$. $$ C D=10 $$, and $$ \angle A B C $$ is a right angle. Find the measare of $$ \overline{A B} $$ in simplified radical form.

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We begin by noting that quadrilateral $$ABCD$$ is inscribed in a circle and the vertices occur in that order. We are given:
- $$ AB = 12 $$
- $$ BC = 16 $$
- $$ CD = 10 $$
- $$ \angle ABC = 90^\circ $$

Since $$\angle ABC$$ is an inscribed angle intercepting arc $$AC$$, and an inscribed angle is a right angle if and only if its intercepted arc is a semicircle, it follows that
$$
AC \text{ is a diameter of the circle.}
$$

In triangle $$ABC$$, with $$\angle ABC = 90^\circ$$, we can use the Pythagorean theorem to find the diameter $$AC$$:
$$
AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20.
$$

Now, since $$AC$$ is a diameter, the angle subtended by $$AC$$ from any point on the circle is a right angle. In particular, in triangle $$ADC$$, $$\angle ADC$$ is a right angle. Applying the Pythagorean theorem in triangle $$ADC$$ (with right angle at $$D$$), we have:
$$
AD^2 + CD^2 = AC^2.
$$
Substitute the known values:
$$
AD^2 + 10^2 = 20^2.
$$
Simplify and solve for $$AD$$:
$$
AD^2 + 100 = 400 \quad \Longrightarrow \quad AD^2 = 400 - 100 = 300.
$$
Taking the square root gives:
$$
AD = \sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3}.
$$

Thus, the measure of $$ \overline{AD} $$ in simplified radical form is 
$$
\boxed{10\sqrt{3}}.
$$
The measure of $$ \overline{AD} $$ is $$ 10\sqrt{3} $$.

15 Quadrilateral ABCD is inscribed in circle \( Q . A B=12 . B C=16 \). \( C D=10 \), and \( \angle A B C \) is a right angle. Find the measare of \( \overline{A B} \) in simplified radical form.

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