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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Solution

Draw perpendicular from $O$ to line $A B$ .
So, $O E ⊥ P Q$ and $O E ⊥ A B$
We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.
So, $P E = E Q$ .....(1)
and $A E = E B$ .....(2)
(2) - (1)
$A E - P E = E B - E Q$
$A P = B Q$
Hence proved

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Similar questions

Q. 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

Q. In the given above figure, there are two concentric circles with centre 'O'. Chord AD of the bigger circle intersects the smaller circle at B and C. Show that AB = CD.

Q. Two equal chord

A B

and

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of a circle with centre

O

, intersect each other at point

P

inside the circle. Prove that:

A P = C P

B P = D P

Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that : $∠ C P A = ∠ D P B$ .

Chords $A B$ and $C D$ of a circle intersect each other at point $P$ such that $A P = C P$ .

Show that : $A B = C D$ .

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

Draw perpendicular from O to line AB. So, OE⊥PQ and OE⊥ABWe know that the perpendicular drawn from the centre of the circle to the chord bisects the chord. So, PE=EQ .....(1) and AE=EB .....(2)(2) - (1)AE−PE=EB−EQAP=BQHence proved

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 $A P = B Q$

Draw perpendicular from $O$ to line $A B$ .
So, $O E ⊥ P Q$ and $O E ⊥ A B$
We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.
So, $P E = E Q$ .....(1)
and $A E = E B$ .....(2)
(2) - (1)
$A E - P E = E B - E Q$
$A P = B Q$
Hence proved