The problem states:

1. The circle is centered at $A$ with a radius of $60$ units.
2. Points $A$, $C$, $D$, and $P$ are collinear.
3. Points $B$ and $D$ are on the circle.
4. The angles $\angle CBD$ and $\angle DBP$ are congruent.
5. We are asked to find the product $AC \times AP$.

Solution Approach

Since $\angle CBD \cong \angle DBP$, point $P$ lies on the tangent from $D$ through $B$. This collinearity and angle equality suggest a geometric relationship involving secant and tangent lines in circle theorems.

Given:

• Radius $OA = 60$ (since $O = A$ is the center).
• $AC \times AP$ can be found using the Power of a Point Theorem, which states that for a point $P$ outside a circle, $AC \times AP = AD \times AB$, where $AB$ and $AD$ are chords passing through point $A$.

Since both $B$ and $D$ are points on the circle, we know:

• $AB = AD =$ the radius, $60$ units.

Thus: $AC \times AP = AB^2 = 60^2 = 3600$

Final Answer

$AC \times AP = 3600$

Would you like further details on this theorem or additional explanations?

Here are some related questions:

1. What is the Power of a Point Theorem and how does it apply to this problem?
2. Can we calculate $AC \times AP$ using other methods?
3. What are secant and tangent lines in relation to a circle?
4. How does the congruence of angles $\angle CBD$ and $\angle DBP$ affect our calculations?
5. Why are $B$ and $D$ on the circle important for this solution?

Tip: Remember, circle theorems like the Power of a Point can simplify complex circle problems.