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 $\angle ADC$.
Show Hint
Tangents drawn at the ends of a diameter are always parallel because the diameter acts as a transversal perpendicular to both lines.
Step 1: Understanding the Concept:
A tangent at any point of a circle is perpendicular to the radius through the point of contact. For lines to be parallel, the sum of interior angles on the same side of the transversal must be $180^\circ$, or alternate interior angles must be equal. Step 2: Key Formula or Approach:
1. Radius $\perp$ Tangent $\implies \angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
2. Sum of angles in a quadrilateral is $360^\circ$. Step 3: Detailed Explanation:
1. Proof of p $\parallel$ q:
- $OA \perp p$ and $OB \perp q$ (Radius is perpendicular to tangent).
- $\angle OAP = 90^\circ$ and $\angle OBQ = 90^\circ$.
- These are alternate interior angles (or consecutive interior angles summing to $180^\circ$).
- Since alternate interior angles are equal, $p \parallel q$.
2. Finding $\angle ADC$:
- In quadrilateral $ABCD$ (trapezium), since $p \parallel q$, then $AD \parallel BC$.
- Consecutive interior angles between parallel lines sum to $180^\circ$.
- $\angle BCD + \angle ADC = 180^\circ$.
- $110^\circ + \angle ADC = 180^\circ$.
- $\angle ADC = 180^\circ - 110^\circ = 70^\circ$. Step 4: Final Answer:
Tangent $p \parallel q$ is proved. The measure of $\angle ADC$ is $70^\circ$.
