Let the two concentric circles have the centre O and let AB be the chord of an outer circle whose length is D and which is also tangent to the inner circle at point D as shown
The diameters are given as d1 and d2 hence the radius will be d1/2 and d2/2
In ΔOAB
⇒ OA = OB …radius of the outer circle
Hence ΔOAB is an isosceles triangle
As radius is perpendicular to tangent OC is perpendicular to AB
OC is altitude from the apex, and in an isosceles triangle, the altitude is also the median
Hence AD = DB = C/2
Consider ΔODB
⇒ ∠ODB = 90° …radius perpendicular to tangent
Using Pythagoras theorem
⇒ OD2 + BD2 = OB2
Multiply the whole by 22
⇒ d22 + C2 = d12
Hence proved