Question

In the adjoining figure, AB is the diameter of the circle with centre O. Two tangents p and q are drawn to the circle at points A and B respectively. Prove that p $\parallel$ q. Further, a line CD touches the circle at E and $\angle BCD = 110^\circ$. Find the measure of $\angle ADC$.

Hint:

Tangents drawn at the ends of a diameter are always parallel because the diameter acts as a transversal perpendicular to both lines.

Answer 1

Concept Used:
The tangent drawn at any point of a circle is always perpendicular to the radius at the point of contact. Also, when a transversal cuts two parallel lines, alternate interior angles are equal and consecutive interior angles sum to $180^\circ$. These properties help in proving parallelism and finding unknown angles.

Proof that $p \parallel q$:
1) Since $OA$ and $OB$ are radii drawn to the points of contact,
$OA \perp p$ and $OB \perp q$.

2) Therefore,
$\angle OAP = 90^\circ$ and $\angle OBQ = 90^\circ$.

3) These right angles are equal. When two lines are perpendicular to the same line (or form equal alternate interior angles), the lines are parallel.

4) Hence, $p \parallel q$.

Finding $\angle ADC$:
1) Since $p \parallel q$, we have $AD \parallel BC$.

2) In trapezium $ABCD$, consecutive interior angles between parallel lines are supplementary.

3) Therefore,
$\angle BCD + \angle ADC = 180^\circ$.

4) Given $\angle BCD = 110^\circ$,
$110^\circ + \angle ADC = 180^\circ$.

5) $\angle ADC = 180^\circ - 110^\circ = 70^\circ$.

Final Result:
The tangents $p$ and $q$ are parallel.
The measure of $\angle ADC$ is $70^\circ$.