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Two circles,centre A and B touch one another at C.Through C a straight line PCQ is drawn cutting the circles at P and Q.Prove that AP || BQ.
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The plot is shown in the Figure.
Draw EF, the straight line perpendicular to AC at the point C.
Then EF is a tangent line to the circle A.
At the same time, DE is the perpendicular to BC at the point C and is a tangent line to the circle B.
So, the line BCA is a straight line.
The angles ACP and BCQ are congruent as they are vertical angles. ( See the lesson (*) )
The angles ECP and FCQ are congruent due to the same reason. ( See the lesson (*) )
Therefore, the minor arcs CP and CQ have the same measures. ( See the lesson (**) )
It implies that the central angles PAC and QBC are congruent.
Thus the triangles ACP and BCQ have two pairs of congruent angles:
(ACP and BCQ, as well as CAP and CBQ).
Hence, the angles APC and BQC are congruent as the pair of the third angles of triangles.
But these angles are alternate interior angles. ( See the lesson (***) )
It implies that the straight line AP is parallel to BQ: AP || BQ. ( See the lesson (***) )
QED.
Solved.
The referenced lessons are
(*) Vertical angles
(**) The angle between a chord and a tangent line to a circle
(***) Parallel lines
Also, you have this free of charge online textbook on Geometry
GEOMETRY - YOUR ONLINE TEXTBOOK
in this site.
The referred lessons are the part if this textbook under the corresponding topics.