Question

5. In figure 6.9, centre of two circles is 0. Chord
AB of bigger circle intersects the smaller circle
in points P and Q. Show that AP=BQ

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Answer 1

Answer:

pls give the figure.....

Answer 2

Given :

• Centre of the circle is O

• AB = Chord

• Chord AB of bigger circle intersects the smaller circle in points P and Q.

To Prove :

• AP = BQ

Construction :

• Draw OM ⊥ AB / PQ

Proof :

$\color{red} \sf \: for \: smaller \: circle$

OM ⊥ PQ

PM = PQ------------(1)

$\sf \color{blue} \: for \: bigger \: circle$

OM ⊥ AB

AM = MB-----------(2)

$\sf AP + \cancel{PM} = \cancel{MQ} + QB$

From (1) = PM = MQ

Therefore, AP = BQ

Hence proved!