Question

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Hint:

When proving tangents equal, always use the congruence of right triangles formed by radii and tangents (RHS rule).

Answer 1

Step 1: Construction.
Let $P$ be a point outside the circle with center $O$. From $P$, draw two tangents $PA$ and $PB$ to the circle, touching it at $A$ and $B$ respectively. Step 2: Join $OA$ and $OB$.
These are radii of the circle. Therefore, \[ OA = OB \] Step 3: Observe right angles.
Since $PA$ and $PB$ are tangents, they are perpendicular to the radii at the point of contact. \[ \angle OAP = \angle OBP = 90^\circ \] Step 4: Consider triangles $\triangle OAP$ and $\triangle OBP$.
In both triangles, \[ OA = OB \quad (\text{radii})
OP = OP \quad (\text{common side})
\angle OAP = \angle OBP = 90^\circ \] Hence, \[ \triangle OAP \cong \triangle OBP \quad \text{(by RHS congruence)} \] Step 5: Conclusion.
By congruence, \[ PA = PB \] Therefore, the lengths of tangents drawn from an external point to a circle are equal.