Let O be the centre of concentric circles with radius ‘r’ and ‘R’ (R>r) and AB be the chord which touches the inner circle at point D
We have to prove that AB has a fixed length
Consider ΔOAB
OA = OB …radius
Hence ΔOAB is an isosceles triangle
Radius OD is perpendicular to tangent AB at the point of contact D
Hence OD is the altitude, and we know that the altitude from the apex of the isosceles triangle is also the median
⇒ AD = BD …(a)
Now consider ΔODB
⇒ ∠ODB = 90° …radius is perpendicular to tangent
Using Pythagoras
⇒ OD2 + BD2 = OB2
The radius are OB = R and OD = r
⇒ r2 + BD2 = R2
⇒ BD2 = R2 - r2
⇒ BD = √(R2 - r2) …(i)
From figure
⇒ AB = AD + BD
Using (a)
⇒ AB = BD + BD
⇒ AB = 2BD
Using (i)
⇒ AB = 2√(R2 - r2)
Here observe that AB only depends on R and r which are fixed radius of inner circle and outer circle.
And as the radius of both the circle will not change however one may draw the chord the radius will always be fixed.
And hence AB won’t change AB is fixed length
Hence proved
Hence, in two concentric circles, all chords of the outer circle which touch the inner circle are of equal length.