Question

Problem

Circles
`K` and
`L` touch externally at point
`P`. Line
`ABC` cuts
`K` at points
`A` and
`B` and is tangent to
`L` at
`C`. The line through
`A` and
`P` meets
`L` at a second point
`D`.

Prove that
`PC` bisects angle
`BPD`.

Answer 1

Explanation:

Using the various theorems relating inscribed and external angles to intercepted arcs, we can write the following relations:

angle A = 1/2(arc BP) . . . . . . . . . . . inscribed angle

angle A = 1/2(arc CD -arc CP) . . . external angle at tangent/secant

angle CPD = 1/2(arc CD) . . . . . . . inscribed angle

angle CPB = 1/2(arc CP) + 1/2(arc BP) . . . . . sum of angles at the mutual tangent

Proof

Equating the expressions for angle A, we have ...

1/2(arc BP) = 1/2(arc CD -arc CP)

Adding arc CP gives ...

1/2(arc BP +arc CP) = 1/2(arc CD)

Substituting the last two equations for angles from above, this gives ...

angle CPB = angle CPD

Hence PC bisects angle BPD.