A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.
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Solution
M is the point of contact of two circles touching externally.
In △APM,AM=AP (radii of G)
∠APM=∠AMP (angle opposite to equal side)
∠AMP=∠QMB (opposite angle) (1)
In △MBQ,MB=BQ=radii
∠QMB=∠BQM (angle opposite to equal side) (2)
From (1) and (2), we get ∠APM=∠BQM
Hence, AP∥BQ as (∠APM,∠BQM are alternate interior angles)