[XY] and [CD] are chords of a circle which, when produced, intersect at a point P outside the circle - Leaving Cert Mathematics - Question 8 - 2010

From the earlier equation, we now have:

Given |XY| + |YP| = |XP| + |PD|, we can substitute: $5 + 4 = |XP| + |PD|\Rightarrow 9 = x + |PD|$

Solving for |PD|, we get: $|PD| = 9 - x$ Since we need to find |PD| explicitly, we also derive it from the product of the segment equation.

To prove this statement, consider the triangle formed by the centre O, point T on the circle, and another point Q on the tangent line. By definition, a tangent line is perpendicular to the radius at the point of contact. Therefore, $|OT| \perp |TQ|$ This also shows that if a line is perpendicular to the radius going to T, it must meet the circle at T only, confirming it's a tangent.

In this circle, with O as the center and |∠PQR| = 42°, we have:

From the property of inscribed angles, it follows that: $|∠PSQ| = \frac{1}{2} |∠PQR| = \frac{1}{2} \times 42° = 21°$

By the supplementary angle property in triangles, we can derive:

Since |∠PQR| = 42° and |∠PQO| is a straight line (180°): $|∠SQP| = 90° - 42° = 48°$

In triangle PQR, we know:

• |∠PQR| = 42°
• |∠SQP| = 48°

Using the angle sum property in triangles: $|∠QPR| = 180° - (|∠PQR| + |∠SQP|) = 180° - (42° + 48°) = 90°$

In triangle QRS, considering the angles:

• |∠PQR| is external to |ΔQRS|,

So: $|∠QRS| = |∠SQP| + |∠QPR| = 48° + 42° = 90°$