Q4:

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Given: A circle with center $O$. Let $AB$ be a diameter of the circle. Let $PQ$ be a tangent at point $A$ and $RS$ be a tangent at point $B$.

To Prove: $PQ \parallel RS$.

Step 1: Applying the Tangent-Radius Theorem
According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."
Since $PQ$ is a tangent at $A$ and $OA$ is the radius, we have:
$OA \perp PQ \implies \angle OAP = 90^\circ$ and $\angle OAQ = 90^\circ$.
Similarly, since $RS$ is a tangent at $B$ and $OB$ is the radius, we have:
$OB \perp RS \implies \angle OBR = 90^\circ$ and $\angle OBS = 90^\circ$.

Step 2: Analyzing the Angles
Consider the lines $PQ$ and $RS$ intersected by the transversal $AB$.
From Step 1, we have:
$\angle OAP = 90^\circ$
$\angle OBS = 90^\circ$
Therefore, $\angle OAP = \angle OBS = 90^\circ$.

Step 3: Establishing Parallelism
Observe that $\angle OAP$ and $\angle OBS$ are alternate interior angles with respect to lines $PQ$ and $RS$ and transversal $AB$.
[Theorem: If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel.]
Since $\angle OAP = \angle OBS = 90^\circ$, the alternate interior angles are equal.
Thus, $PQ \parallel RS$.

Final Answer: Since the alternate interior angles formed by the transversal $AB$ with lines $PQ$ and $RS$ are equal ($90^\circ$), the tangents $PQ$ and $RS$ are parallel to each other.