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
- <OEA + <OEA + <AED = 180° (Just put the values) ..(1)
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:
- <OEA + <OED + <AED = 180°
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`.
