Question

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

Answer 1

When you draw a perpendicular (OM) from centre o to chord AB, it bisect the chordAB such that AM =MB. Now, this perpendicular is also bisecting the smaller chord PQ such that PM=MQ. Now, subtracting PM=MQ from AM =MB.
AM - PM =MB - MQ
AP = BQ
HENCE, PROVED

Answer 2

hence it proved........

hope it helps you ...........