Step 1: Understand the problem.
A circle with center $O$ has tangents $PA$ and $PB$ drawn from an external point $P$ to the points of tangency $A$ and $B$. A third tangent at point $E$ intersects $PA$ at $C$ and $PB$ at $D$. Given $PA = 10$ cm, find the perimeter of $\triangle PCD$.
Step 2: Tangent property.
Tangents from an external point to a circle are equal in length. Therefore, $PA = PB = 10$ cm.
Step 3: Tangent segments from $C$ and $D$.
Let $E$ be a point on the circle where a tangent line intersects $PA$ at $C$ and $PB$ at $D$.
Since $C$ is an external point to the circle, $CA = CE$.
Since $D$ is an external point to the circle, $DB = DE$.
Step 4: Calculate the perimeter of $\triangle PCD$.
The perimeter of $\triangle PCD$ is $PC + CD + PD$.
We can express $CD$ as $CE + ED$.
Substituting $CE = CA$ and $ED = DB$, we get $CD = CA + DB$.
So, the perimeter is $PC + CA + DB + PD$.
Notice that $PC + CA = PA$ and $PD + DB = PB$.
Therefore, the perimeter of $\triangle PCD = PA + PB$.
Given $PA = 10$ cm and $PB = 10$ cm.
Perimeter of $\triangle PCD = 10 \, \text{cm} + 10 \, \text{cm} = 20 \, \text{cm}$.
Conclusion:
The perimeter of triangle $PCD$ is $20$ cm. The power of a point theorem was not directly needed for this calculation, but the property of equal tangents from an external point was crucial. The key insight is that the segments of the third tangent ($CE$ and $DE$) can be related to segments of the original tangents ($CA$ and $DB$), which simplifies the perimeter calculation.