In the adjoining figure, $O$ is the centre of a circle. If $A B$ and $A C$ are chords of the circle such that $A B = A C, O P ⊥ A B$ and $O Q ⊥ A C$ , prove that $P B = Q C$

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Solution

It is given that $A B = A C$

Dividing the equation by $2$

We get

$\frac{1}{2} A B = \frac{1}{2} A C$

Perpendicular from the centre of a circle to a chord bisect the chord
$M B = N C . . . (1)$

We know that the equal chords are equidistant from the centre if the circle
$O M = O N$ and $O P = O Q$

Subtracting both the equation

$O P - O M = O Q - O N$

So we get

$P M = Q N . . . . . (2)$

Consider $△ M P B$ and $△ N Q C$

We know that

$∠ P M B = ∠ Q N C = 90^{o}$

By $S A S$ congruence criterion

$△ M P B ≅ △ N Q C$

$P B = Q C$ (c.p.c.t)

Therefore, it is proved that $P B = Q C$

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Similar questions

In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, OP $⊥$ AB and OQ $⊥$ AC, prove that PB = QC.

In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.

Q. In the adjoining figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show that AC || CD and AC = 2 × OD.

Q. In the given figure, AB and AC are equal chords of a circle with centre O and

O P ⊥ A B, O Q ⊥ A C

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The relation between

P B

Q C

In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisector of $∠$ BAC.

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In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB=AC,OP⊥AB and OQ⊥AC, prove that PB=QC

In the adjoining figure, $O$ is the centre of a circle. If $A B$ and $A C$ are chords of the circle such that $A B = A C, O P ⊥ A B$ and $O Q ⊥ A C$ , prove that $P B = Q C$