# Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A

**Solution:**

Given, AB is the __diameter of a circle__.

AB bisects all those chords which are parallel to the tangent at the point A.

From the figure,

AB is the diameter of the circle.

CD is the __chord of the circle__.

MN is the tangent to the circle at the point A.

OA = OB = OC = OD = radius of the circle.

We know that the __radius of a circle__ is perpendicular to the tangent at the point of contact.

So, OA ⟂ MN

Also, ∠MAO = ∠NAO = 90°

The __corresponding angles__ are equal.

∠CEO = ∠MAO

So, ∠CEO = 90°

We know that the __perpendicular__ drawn from the centre of the circle to the chord always bisects the chord.

It implies that OE bisects CD.

Therefore, diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

**✦ Try This: **Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.

**☛ Also Check:** NCERT Solutions for Class 10 Maths Chapter 10

**NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 10**

## Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A

**Summary:**

It is proven that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A

**☛ Related Questions:**

- In Fig. 9.15, from an external point P, a tangent PT and a line segment PAB is drawn to a circle wit . . . .
- If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, resp . . . .
- If a hexagon ABCDEF circumscribes a circle, prove that AB + CD + EF = BC + DE + FA

visual curriculum