Prove that the tangent drawn at the mid - point of an arc of a circle is parallel to the chord joining the end point of the arc.

Given : An arc ABE of a circle with center O . B is a point on arc ABE such that arc AB = arc BC and DE is a tangent at point B

To prove : chord AB is parallel to Tangent DE

Proof :

As arcs AB and BC are of equal lengths. And lies in same circle. So areas of their respective sector will be equal

And in case of sectors

i.e. the area is only dependent on angle because π and r are constants

so if areas are equal then angles will also be equal .

i.e. ∠1 = ∠2

In △AOF and △FOC

AO = OC [radii of same circle]

∠1 = ∠2 [Proved above]

OF = OF [Common]

△AOF ≅ △FOC [BY SAS CRITERION]

∠OFA = ∠OFC [By CPCT]

But

∠OFA + ∠OFC = 180°

∠OFA + ∠OFA = 180°

∠OFA = 90° [1]

Also,

OB⏊DE

∠DBF = 90° [2]

[tangent drawn at any point to a circle is perpendicular to the radius through the point of contact]

As we know two lines are parallel if their corresponding angles are equal.

From 1 and 2

AB is parallel to DE .