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.

..

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.

..