# In figure 6.9, center of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ

We draw a perpendicular on chord AB from O.

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

Therefore,

AM = MB …….(1)

OM is also perpendicular to chord PQ of smaller circle.

Therefore,

PM = MQ ………….(2)

Subtracting (2) from (1)

AM-PM = MB-MQ

AP = BQ

Hence Proved.

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