Two circles C1 and C2 touch externally at A. The tangent to C1 drawn at a point B on the circumference of C1 meets C2 in P and Q so that P is between B and Q. QA is produced and meets C1 at X. Prove that angle XAB = angle PAB.

To prove that angle XAB is equal to angle PAB, we can use some properties of tangent lines and angles formed between intersecting lines and circles.

Let's break down the problem step by step:

Step 1: Draw a clear diagram of the given information. Label the points as mentioned in the question: A, B, P, Q, X.

Step 2: Identify the angles we need to compare: angle XAB and angle PAB.

Step 3: Since we have tangents drawn from point B to C1, we know that the angle formed between the tangent line and the radius at the point of tangency (B) is 90 degrees. Therefore, angle BAC is 90 degrees.

Step 4: Now, let's look at triangle ABX. Since BX is a tangent to circle C1, angle XAB is equal to the angle formed between the tangent line (BX) and the radius (AB) at point B.

Step 5: Since P and Q are points of intersection between the tangent line (BX) and C2, we can use the property that the angle formed between a tangent and a chord of a circle is equal to the angle formed between the chord and the radius at the point of intersection. In this case, angle PBQ is equal to the angle formed between the chord (PQ) and the radius (AB) at point P.

Step 6: Now, focusing on triangle ABP, we see that angle PAB is equal to the angle formed between the chord (PQ) and the radius (AB) at point P.

Step 7: Combining the information from steps 4, 5, and 6, we can conclude that angle XAB is equal to angle PAB.

Therefore, we have proved that angle XAB is equal to angle PAB.

Remember, when working on geometry proofs, it's crucial to draw accurate diagrams and use the properties and theorems related to the given shapes and angles.