2 circles C(O,r) and C( O',r') intersect at 2 points A and B and O lies on C(O',r'). A tangent CD is drawn to the circle C(O',r') at A.Then
a. angle(OAC)= OAB
b. angle( OAB)=AO'O
c. angle (AO'B)=AOB
Also triangles AOO' and BOO' are identical.
so angle O'AO=AOO'=OBO'=BOO' = x
So OO'A = OO'B = 180-2x
angle O'AC=90 (tangent + radius)
So OAB= 90-x
Let AB and OO' intersect at E.
In triangle OAE, angle OEA=90.
Since <EOA = x, angle OAE=OAB=90-x
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