Question: Two parallel tangents to a circle, centre O, are cut a P and Q by a third tangent.
Find the size of angle POQ.
I have drawn the diagram and used theorems like tangents cut at external point; radius-tangent theorem. But this is not working.
I really appreciate it if someone could explain this question to me.
Use the diagram, where i constructed a few lines like the radius to tangents OA OB OC:
http://tinypic.com/r/2ecjkeh/4
Let angle AOP=ø
First step, OAP=90degrees (because line from centre to tangent is perpendicular to tangent)
Same for OCP=90 degrees
Hence, you can say that OAPC is a cyclic quadrilateral (opposite sides are supplementary)
and because OAPC is cyclic quad, then 2ø = 180- angleAPC (opp sides of cyclic quad)
which can be rewritten as 2ø = CQB or 2ø = CQO+OQB (cointerior angles of parallel lines are supplementary, therefore 180-APC=CQB)
Then you can prove that triangle OCQ is congruent to triangle OBQ using RHS with common hypotenuse, radii and 90 degrees created from line from centre to tangent, thus COQ=BOQ and CQO=OQB
COQ+CQO=90 (angle sum of triangle =180, where angle OCQ=90 because line from centre to tangent=90)
Which can be rewritten as ø+ COQ = 90...which is also the same as POQ
Therefore POQ =90 degrees.
Sorry if it is confusing, but really hard to explain circle geo questions in this way. Probably an easier way.
