two circles c1 and c2 meet at the points A and B. CD is a common tangent to these circles where C and D lie on the circumference of C1 and C2 respectively. CA is the tangent to c2 at A. When produced DB meets the circumference of C1 at p. Prove that PC is parallel to AD
To prove that PC is parallel to AD, we can use the concept of angles formed by intersecting lines and tangents to circles.
Let's start by considering triangle CAD. Since CA is a tangent to c2 at A, the angle CAD is a right angle. So, we have a right triangle CAD.
Next, let's consider triangle BDA. We know that DB intersects the circumference of c1 at point P. Therefore, angle PDB is an angle formed by the tangent CD and the chord PB. By the tangent-chord angle theorem, angle PDB is equal to the angle formed between the tangent CD and the chord PB at the point of intersection (D in this case). That is, angle PDB = angle DCP.
Now, let's consider triangle CPD. Since PDB and DCP are equal, the sum of their angles (angle PCD) must be equal to 180 degrees. In other words, angle PCD is a straight angle.
Since angle PCD is a straight angle, we can conclude that PC and AD are parallel. This is because when a pair of lines intersect, the opposite angles they form are equal (or supplementary if they form a straight angle). In this case, angle PCD and angle CAD are opposite angles formed by the intersection of lines PC and AD, and since angle PCD is a straight angle, angle CAD must also be a straight angle (180 degrees), making PC parallel to AD.
Therefore, we have proved that PC is parallel to AD.