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Let ABC is the given ISOSCELES triangle.  Where angle B=80, C=80,A=20 respectively. (in degrees)

We need to find thae angle ADC where D is a point on AB such that AD = BC.
(Here after let us refer isosceles triangle is "iso")

SOLUTION:
 
Draw a line CE such that E is on AB and CE = BC. i,e, BCE is isosceles.
 
Observations: (all angles in degrees)
EBC= 80  , BEC = 80 (as triangle BCE is iso ) , BCE = 20 ( from triangle sum = 180)
ECF = BCF - BCE = 80 - 20 = 60

Draw a line EF such that F is on AC and EF = EC (which is = BC) . i,e, CEF is isosceles.
 
Observations: (all angles in degrees)
ECF= 60  , CFE = 60 (as triangle CEF is iso ) , CEF = 60 ( from triangle sum = 180)
From above all angles are 60. So CEF is equilateral.
So, FC = BC
GEF = 180 - ( BEC + CEF)  = 40

Draw a line FG such that G is on AB and FG = EF (which is = BC) . i,e, EFG is isosceles.
 
Observations: (all angles in degrees)
GEF= 40  , EGF = 40 (as triangle EFG is iso ) , EFG = 100 ( from triangle sum = 180)
AGF = 180 - EGF  = 180 - 40 = 140
AFG =  180 - ( CFE + EFG) = 20
We can see that GAF = AFG = 20 
                 so AGF is iso and AG = FG . But FG = BC 
So AG = BC. 
So G is our required point.

In triangle GFC, FG = FC =BC from above observations. hence GFE is iso. 
       so, angle FGE = FCE = 10 , (as GFE = 160 and sum of angles in triangle = 180)
Angle AGC = AGF + FGC 
   = 140 + 10 
=150 degrees.

P.S. : I observed it with different angles and found that there is unique answer for each angle of isosceles triangle. 
  Here angle A = 20, answer is ADC = 150
    A = 0   ->  ADC = 180 
    A = 20   ->  ADC = 150
    A = 40  ->  ADC = 100
    A = 60   ->  ADC = 60
    A = 80   ->  ADC = 40
         ........
If anyone can derive a generalised theorem to find it, please post it here or mail it.
..

 
 
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Definition:
A chord is the line joining any 2 points on the circle.
         In the figure, AB,CD,EF are chords.
Property 1:
The line joining the midpoint of a chord and the centre of the circle is perpendicular to the chord.
                  OR
The perpendicular bisector of any chord passes through the center.
                 OR
The perpendicular dropped from the centre of the circle to any chord will bisect the chord.

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The mid points of all the chords of equal length will form a circle with the center same as the original circle.

Distance of the chord from the centre of the circle = root(r*r - s*s/4)