In figure above, centre of two circles is $O$ . Chord $A B$ of bigger circle intersects the smaller circle in points $P$ and $Q$ Show that $A P = B Q$

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Solution

As perpendicular from centre bisects the chord
$\Rightarrow P M = M Q$ __(1)
and $A M = M B$
$\Rightarrow A P + P M = M Q + Q B$
$\Rightarrow A P = Q B$ (By (1))

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In the given figure, centre of two circles is $O$ . Chord $A B$ of bigger circle intersects the smaller circle in points $P$ and $Q$ . Show that $A P = B Q$

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

As perpendicular from centre bisects the chord ⇒PM=MQ __(1)and AM=MB⇒AP+PM=MQ+QB⇒AP=QB (By (1))

In figure above, centre of two circles is $O$ . Chord $A B$ of bigger circle intersects the smaller circle in points $P$ and $Q$ Show that $A P = B Q$

As perpendicular from centre bisects the chord
$\Rightarrow P M = M Q$ __(1)
and $A M = M B$
$\Rightarrow A P + P M = M Q + Q B$
$\Rightarrow A P = Q B$ (By (1))