prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given : A circle C (0, r) and a tangent l at point A.
To prove : OA ⊥ l
Construction : Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C.
Proof: We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l.
OA = OC (Radius of the same circle)
Now, OB = OC + BC.
∴ OB > OC
⇒ OB > OA
⇒ OA < OB
B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line segment joining O to any point on l.
Here, OA ⊥ l