In \(\Delta ABC\), given \(AC = BC\) and \(DE\) is the diameter of the circle. \(AC\) and \(BC\) touch the circle at points \(M\) and \(N\), respectively. If \( \angle ADP = \angle BEQ = 100^\circ \), find \( \angle PRD \).
Show Hint
In geometric problems involving circles and tangents, always use the property that an angle subtended by the diameter is a right angle. Additionally, remember that vertically opposite angles are equal, which is very useful when solving for unknown angles.
Step 1: Understanding the given information We are given an isosceles triangle \( \Delta ABC \) where \( AC = BC \) and \( DE \) is the diameter of the circle. The tangents at points \(M\) and \(N\) are drawn from \(A\) and \(B\) to the circle. The angles \( \angle ADP = \angle BEQ = 100^\circ \).
Step 2: Property of angles subtended by the diameter Since \(DE\) is the diameter of the circle, the angle subtended by the diameter at any point on the circle is a right angle. Hence:
\[
\angle DMN = 90^\circ
\]
Step 3: Relationship between angles in triangle ADP We are given that \( \angle ADP = 100^\circ \). In \( \triangle ADP \), the angle sum of a triangle is \(180^\circ\), so:
\[
\angle PDE = 180^\circ - 100^\circ = 80^\circ
\]
Step 4: Calculate \( \angle PED \) In triangle \( \triangle APE \), the sum of angles is \(180^\circ\). Therefore:
\[
\angle PED = 180^\circ - 90^\circ - 80^\circ = 10^\circ
\]
Step 5: Calculate \( \angle QER \) in triangle QRE Next, in \( \triangle QRE \), we can use \( \angle PED = 10^\circ \) to find:
\[
\angle QER = 80^\circ - 10^\circ = 70^\circ
\]
Step 6: Vertical angles in \( \Delta QRE \) and \( \Delta PRD \) Since \( \angle QRE = \angle PRD \) (they are vertically opposite angles), we conclude:
\[
\angle PRD = 20^\circ
\]
Final Answer: The correct answer is (a) 20°.
