how alternate segment theorem can be prove

Hi!

 

Alternate segment theorem can be stated as “If a line touches a circle and a chord is drawn from the point of contact, then the angle between the tangent and the chord are respectively equal to the angles in the corresponding alternate segments”

 

In the above figure TPT’ is a tangent. PR is a chord through P. According to alternate segment theorem ∠RPT = ∠RSP

 

Let PO is extended. Let it intersects the circle at point U. Join U to R.

 

POU is the diameter of the circle.

Thus, POURQP is a semicircle.

∴ ∠URP = 90°

 

Using angle sum property for triangle UPR

∠URP + ∠RUP + ∠UPR = 180°

⇒ 90° + ∠RUP + ∠UPR = 180°

⇒ ∠RUP + ∠UPR = 90°  … (1)

 

TPT’ is a tangent and OP is the radius.

∴ ∠UPT = 90°

⇒ ∠RPT + ∠UPR = 90°  … (2)

 

From equations (1) and (2)

∠RUP = ∠RPT   … (3)

 

But ∠RUP and ∠RSP  are angles in the segment PQR

∠RUP = ∠RSP   … (4)

 

From (3) and (4)

∠RPT = ∠RSP

 

Hope! You got the concept of proving alternate segment theorem.

 

Cheers!