In two concentric circles, prove that all chords of the outer circle which touch the inner arc of equal length.

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.