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Question

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, APBQ as (APM,BQM are alternate interior angles)

894672_564060_ans_e83340a4c34f4bb0ac8e80bd575fa3dd.png

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