If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle.
Proof
Let DE touch the circle at B, and BC be a chord from B into
the circle. We show ∠DBC equals the inscribed angle in
the alternate segment BAC.
Draw the diameter BA from B. By III.18, the tangent DE is
perpendicular to BA, so ∠DBA= right. By III.31, the
inscribed angle ∠BCA in the semicircle is right. In
△BCA, by I.32, ∠BCA+∠CAB+∠ABC= two right angles; since ∠BCA is right, ∠CAB+∠ABC= one right angle.
Now ∠DBC=∠DBA−∠ABC= right −∠ABC=∠CAB (from the identity above). By III.21, ∠CAB
equals any other inscribed angle in the same segment as A.
Hence ∠DBC equals the inscribed angle in the alternate
segment. The other angle ∠EBC on the tangent's other side
equals the inscribed angle in the original segment by analogous
argument.