In the adjoining figure, AB is a chord of a circle with centre O and BC is a diameter. If OD ⟂ AB, show that CA = 2OD and CA ∥ OD.
Circles
2 Likes

Question

In the adjoining figure, AB is a chord of a circle with centre O and BC is a diameter. If OD ⟂ AB, show that CA = 2OD and CA ∥ OD.

KnowledgeBoat · Accepted Answer

Since, the perpendicular to a chord from the centre of the circle bisects the chord, ∴ AD = DB We can say that D is mid-point of AB. Since, BC is diameter and O is center so, OB = OC = radius. We can say that O is mid-point of BC. In △ABC, Since, D is mid-point of AB and O is mid-point of BC. By mid-point theorem, ⇒ OD || AC and OD = AC ⇒ AC = 2OD. Hence, proved that CA = 2OD and CA ∥ OD.

Mathematics

In the adjoining figure, AB is a chord of a circle with centre O and BC is a diameter. If OD ⟂ AB, show that CA = 2OD and CA ∥ OD.

Circles

Related Questions

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.

The adjoining figure shows a circle with centre O in which a diameter AB bisects the chord PQ at point R. If PR = RQ = 8 cm and RB = 4 cm, find the radius of the circle.

In the adjoining figure, P is a point of intersection of two circles with centres C and D. If the straight line APB is parallel to CD, prove that AB = 2CD.

If a diameter of a circle bisects each of the two chords of a circle, then prove that the chords are parallel.