A cyclic quadrilateral ABCD has a tangent line l to its circumcircle at A. Line l meets line CD at P, with AD = 3, DP = 8 and PA = 10. Find the length of CD.

Step-by-step solution

1. 1

   Identify the configuration at P. From the external point $P$ there are two lines to the circle: the tangent $PA$, touching at $A$, and the secant through $D$ and $C$. That is exactly the setting for the tangent–secant (power of a point) theorem.

2. 2

   State the theorem. The power of $P$ with respect to the circle can be computed two ways, giving

   $PA^{2}=PD\cdot PC$

   where $PD$ and $PC$ are the distances from $P$ to the near and far intersection points of the secant. Note the tangent length is squared, and the secant contributes the whole product — not $PD\cdot DC$.

3. 3

   Substitute the given lengths.

   $10^{2}=8\cdot PC\quad\Longrightarrow\quad 100=8\,PC\quad\Longrightarrow\quad PC=\frac{100}{8}=\frac{25}{2}=12.5$

4. 4

   Subtract to get the chord. Since $D$ lies between $P$ and $C$ on the secant,

   $CD=PC-PD=\frac{25}{2}-8=\frac{25}{2}-\frac{16}{2}=\frac92=4.5$

   $\boxed{CD=\dfrac92}$

5. 5

   Check consistency. With $PD=8<PC=12.5$, the point $D$ is indeed the nearer intersection, as assumed. Also $PA=10$ lies strictly between $PD=8$ and $PC=12.5$, which is forced by the theorem: $PA=\sqrt{PD\cdot PC}$ is the geometric mean of the two, and a geometric mean always sits between the two numbers. The datum $AD=3$ is not needed for this part — it belongs to the later parts of the question.

6. 6

   Note the triangle inequality as a further sanity check. In triangle $APD$ the sides are $PA=10$, $PD=8$ and $AD=3$: $8+3=11>10$ ✓, so such a triangle exists and the configuration is consistent.