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

Let there be two circles with center at O. Let AB and CD be two chords of the outer circle which touch the inner circle at M and N respectively.

Let radius of outer circle and inner circle be R and r respectively.

AB is a tangent to the inner circle.

∴ OM ⊥ AB

In right angle triangle OMB,

⇒ OB² = OM² + MB² [By pythagoras theorem]

⇒ R² = r² + MB²

⇒ MB² = R² - r²

⇒ MB = $\sqrt{R^2 - r^2}$ ………..(1)

As, perpendicular from center bisects chord.

∴ MB = $\dfrac{AB}{2}$

Substituting value of MB in (1), we get :

$\Rightarrow \dfrac{AB}{2} = \sqrt{R^2 - r^2} \\[1em] \Rightarrow AB = 2\sqrt{R^2 - r^2} ………….(2)$

CD is a tangent to the inner circle.

∴ ON ⊥ CD

In right angle triangle OND,

⇒ OD² = ON² + ND² [By pythagoras theorem]

⇒ R² = r² + ND²

⇒ ND² = R² - r²

⇒ ND = $\sqrt{R^2 - r^2}$ ………..(3)

As, perpendicular from center bisects chord.

∴ ND = $\dfrac{CD}{2}$

Substituting value of ND in (3), we get :

$\Rightarrow \dfrac{CD}{2} = \sqrt{R^2 - r^2} \\[1em] \Rightarrow CD = 2\sqrt{R^2 - r^2} …………..(4)$

From (3) and (4), we get :

AB = CD.

Hence, proved that all chords of the outer circle, which touch the inner circle, are of equal length.