In the given figure, common tangents, AB and CD to two circles with centres O and O', intersect at E. Prove that the points O, E, O' are collinear.

Question

contestada

Respuesta :

MisterBrian MisterBrian · Answer 1 · 2024-01-24T21:59:36+01:00

Answer:

Given below:

Step-by-step explanation:

Common tangents are those tangents that touch on more than one circle.

Since AE and CE are two tangents originating from point E exterior of circle, so they are equal.

Construction: Draw OA,OC,OD,OB as radius.

In ∆ AEO and ∆ CEO

AE = CE (See above)

OA.= OC ( Radius)

OE = OE (Common)

By SAS congruency, ∆AEO is congruent to ∆ CEO

By c.p.c.t.c. <AEO = <CEO

As CD is a common tangent,it must be a straight line

We can now apply linear pair theorem in which the sum of angles on a line divided by any number of line equals 180°

But <CEO = < OEA(See Above)

But < OED = < OEA + <AED (From figure)

So

In ∆ OEC and ∆ OED,

<OEC = < OED( Vertically opposite angles)

But <OEC = <OEA

Hence <OEA = <OED

So we can replace <OEA with <OED in eqn 1:

So we see OEO` is a line in which the sum of angles divided by lines AE and DE equals 180°,so we conclude O,E,O` are collinear points on OEO`.