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.

Open in App

Solution

Given: AB and AC are chords of the circle with centre O. AB = AC, OP ⊥ AB and OQ ⊥ AC

To prove: PB = QC
Proof:
AB = AC (Given)
⇒ $\frac{1}{2} AB = \frac{1}{2} AC$
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ MB = NC ...(i)
Also, OM = ON (Equal chords of a circle are equidistant from the centre)
and OP = OQ (Radii)
⇒ OP - OM = OQ - ON
∴ PM = QN ...(ii)
Now, in ΔMPB and ΔNQC, we have:
MB = NC [From (i)]
∠PMB = ∠QNC [90° each]
PM = QN [From (ii)]
i.e., ΔMPB ≅ ΔNQC (SAS criterion)
∴ PB = QC (CPCT)

Suggest Corrections

Similar questions

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

.
The relation between

P B

Q C

Q. O is the centre of a circle in which AB and AC are congruent chords. Radius OP is perpendicular to chord AB and radius OQ is perpendicular to chord AC. If

∠ P B A = 30^{\circ}

, show that PB is parallel to 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.

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.

AB and AC are two chords of a circle of radius $r$ such that $A B = 2 A C$ . If $p$ and $q$ are the distances of $A B$ and $A C$ from the centre then prove that $4 q^{2} = 3 r^{2} + p^{2}$

Join BYJU'S Learning Program