Question

Prove that tangent segments drawn from an external point to a circle are congruent.

Hint:

This is a standard theorem proof. The key steps are always: 1. Draw radii to the points of contact. 2. Identify the right angles using the tangent-radius theorem. 3. Prove the two triangles formed are congruent using RHS. 4. Conclude the tangent segments are equal by CPCTC (Corresponding Parts of Congruent Triangles).

Answer 1

Step 1: Understanding the Concept: 
We need to prove a fundamental theorem of circles which states that if we take a point outside a circle and draw two lines from it that are tangent to the circle, the lengths of these tangent segments (from the external point to the point of tangency) are equal. 
 

Step 2: Key Formula or Approach: 
The proof involves using the properties of tangents and radii, and proving the congruence of two right-angled triangles. We will use the Right-angle-Hypotenuse-Side (RHS) congruence criterion. 
 

Step 3: Detailed Explanation: 
Given: \[\begin{array}{rl} \bullet & \text{A circle with centre O.} \\ \bullet & \text{An external point P.} \\ \bullet & \text{PA and PB are two tangent segments from P to the circle at points A and B respectively.} \\ \end{array}\] To Prove: \[ \text{seg } PA \cong \text{seg } PB \text{ (or } PA = PB \text{)} \] Construction: Draw segments OA, OB, and OP. 
Proof: Consider \(\triangle\)OAP and \(\triangle\)OBP. \[\begin{array}{rl} 1. & \text{\(\angle OAP = \angle OBP = 90^\circ\) \dots(Tangent-radius theorem states that the radius is perpendicular to the tangent at the point of contact)} \\ 2. & \text{seg OA \(\cong\) seg OB \dots(Radii of the same circle)} \\ 3. & \text{seg OP \(\cong\) seg OP \dots(Common side)} \\ \end{array}\] Therefore, by the Right-angle-Hypotenuse-Side (RHS) congruence rule, \[ \triangle OAP \cong \triangle OBP \] Since the triangles are congruent, their corresponding sides must be equal. \[ \therefore \text{seg } PA \cong \text{seg } PB \]

Step 4: Final Answer: 
Hence, it is proved that tangent segments drawn from an external point to a circle are congruent.