Question

prove that the line segment joining the point of contact of two parellel tangents to a circle is a diametre of the circle.​

Answer 1

proof is below check it

Done

Answer 2

Given :

CD and EF are two parallel tangents at the points A and B of a circle with centre O.

To prove :

AOB is the diameter of the circle.

Construction :

Join OA and OB. Draw OG||CD.

Proof :

OG||CD and AO cuts them.

∴ ∠CAO + ∠GOA = ${\sf {180}^{\circ}}$

We know that, OA⏊CD

$\sf \dashrightarrow \angle{{90}^{\circ} + \angle{GOA} = {180}^{\circ}}$

${\sf \dashrightarrow \angle{GOA} = {90}^{\circ}}$

Similarly,

$\sf \dashrightarrow \angle{GOB} = {90}^{\circ}$

$\sf \dashrightarrow \angle{GOA} + \angle{GOB} = {90}^{\circ} + {90}^{\circ} = {180}^{\circ}$

So,

$\sf \dashrightarrow AOB \: is \: a \: straight \: line$

Thus, AOB is a diameter of the circle with centre O.