Theorem 141: The tungent at any point of a circle isradias chrougt the

1.	Theorem 141: The tungent at any point of a circle isradias chrougt the peoint of contactperpendiculer4. Drawothe
Answer» Given :A circle C (0,r) and a tangentlat point A. To prove :OA⊥l Construction :Take a point B, other than A, on the tangentl. Join OB. Suppose OB meets the circle in C. Proof:We know that, among all line segment joining the point O to a point onl, the perpendicular is shortest tol. OA = OC (Radius of the same circle) Now, OB = OC + BC. ∴ OB > OC ⇒ OB > OA ⇒ OA < OB B is an arbitrary point on the tangentl. Thus, OA is shorter than any other line segment joining O to any point onl. Here, OA ⊥l

Theorem 141: The tungent at any point of a circle isradias chrougt the peoint of contactperpendiculer4. Drawothe

Answer»

Given :A circle C (0,r) and a tangentlat point A.

To prove :OA⊥l

Construction :Take a point B, other than A, on the tangentl. Join OB. Suppose OB meets the circle in C.

Proof:We know that, among all line segment joining the point O to a point onl, the perpendicular is shortest tol.

OA = OC (Radius of the same circle)

Now, OB = OC + BC.

∴ OB > OC

⇒ OB > OA

⇒ OA < OB

B is an arbitrary point on the tangentl. Thus, OA is shorter than any other line segment joining O to any point onl.

Here, OA ⊥l