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

Given : Two concentric circles C₁ and C₂. AB and CD are the chords of the outer circle C₁, such that they touch the inner circle C₂ at E and F respectively.
Join OA, OB, OC, OD, OE and OF. ∵ OE ⊥ AB
[Tangent at any point of circle is perpendicular to the radius through the point of
contact]
⇒    ∠OEA = 90°
Similarly, ∠OFC = 90°
∴ ∠OEA = ∠OFC    ...(i)
Now, in right ΔAOE and ΔCOF
OA = OC (radii of the circle) OE = OF (radii of the circle)
and ∠OEA = ∠OFC = 90° [from ...(i)]
Using R.H.S. condition, we get ΔAOE = ΔCOF
∴ EA = FC    ...(i)
Similarly, we can prove
EB = FD    ....(ii)
Adding (i) and (ii), we get
EA + EB = FC + FD) AB = CD.