Problem Analysis:
A circle with center \( O \) has tangents \( PA \) and \( PB \) drawn from an external point \( P \), where \( A \) and \( B \) are the points of tangency. A third tangent is drawn at point \( E \) on the circle, intersecting \( PA \) at \( C \) and \( PB \) at \( D \). Given \( PA = 10 \, \text{cm} \), the objective is to find the perimeter of \( \triangle PCD \).
Application of Tangent Properties:
From the property that tangents from an external point to a circle are equal in length, we have \( PA = PB = 10 \, \text{cm} \).
Also, \( CA \) and \( CE \) are tangents from \( C \) to the circle, so \( CA = CE \).
Similarly, \( DB \) and \( DE \) are tangents from \( D \) to the circle, so \( DB = DE \).
Calculating Perimeter of \( \triangle PCD \):
The perimeter of \( \triangle PCD \) is \( PC + CD + PD \).
We can express \( PC \) as \( PA - CA \) and \( PD \) as \( PB - DB \).
Substituting \( PA = 10 \) and \( PB = 10 \), we get \( PC = 10 - CA \) and \( PD = 10 - DB \).
The segment \( CD \) can be expressed as \( CE + DE \).
Substituting the equal tangent lengths, \( CD = CA + DB \).
Now, the perimeter of \( \triangle PCD \) is:
\( \text{Perimeter} = PC + CD + PD \)
\( \text{Perimeter} = (10 - CA) + (CA + DB) + (10 - DB) \)
\( \text{Perimeter} = 10 - CA + CA + DB + 10 - DB \)
\( \text{Perimeter} = 10 + 10 \)
\( \text{Perimeter} = 20 \, \text{cm} \)
Conclusion:
The perimeter of \( \triangle PCD \) is \( 20 \, \text{cm} \). This result is derived using the properties of tangents from an external point and is independent of the position of point \( E \) or the lengths of \( PC \) and \( PD \) individually. The key insight is that \( CD = CE + DE = CA + DB \), which simplifies the perimeter calculation.