# Alternate segment theorem

#### Summary*arrow_upward*

A line that touches a circle at only one point is known as a tangent to the circle.

The common point to the tangent and the circle is known as the point of contact.

**Alternate segment theorem
**

If a line touches a circle and from the point of contact a chord is drawn, then the angles that this chord makes with the given line are equal to the angles formed in the corresponding alternate segments, respectively.

Given: Let AB be a chord of a circle with centre O. PQ be a tangent to the circle at A.

Let E and F be any two points on the circle such that they are in alternate segments R2 and R1, respectively.

To prove: (i) m BAQ = m AEB

(ii) m BAP = m AFB

Proof:

AEB is an inscribed angle in arc AEB, while arc AFB is the intercepted arc.

m AEB = (1/2)m( arc AFB) (By inscribed angle theorem) ….. (1)

PQ is tangent at A and line AB is a secant.

BAQ intercepts arc AFB

m BAQ = (1/2)m( arc AFB) (By tangent secant theorem) ….. (2)

From (1) and (2)

m AEB = m BAQ

Similarly, m AFB = m BAP.

Converse of Alternate Segment Theorem

If a line is drawn through an end point of a chord of a circle so that the angle formed with the chord is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle.

Given: AB is a chord of a circle with centre O.

Line PAQ is drawn through A such that m BAQ = m ACB , where C is a point on the circumference.

To prove : PAQ is a tangent to the circle at point A.

Proof:

Let us suppose that PAQ is not a tangent.

Draw a tangent P'AQ' to the circle at A.

m BAQ' = m ACB (By alternate segment theorem)

But m BAQ = m ACB (Given)

m BAQ = m BAQ'

Unless ray AQ' coincides with AQ, this is impossible.

Therefore, P'AQ' coincides with PAQ.

Or PAQ is a tangent to the circle at A.

#### Questions & Answers*arrow_upward*

### 1 . Prove that any four vertices of a regular pentagon are concyclic

Given ABCDE is a regular pentagon

That is AB = BC = CD = DE = AE

Recall that the sum of angles in a regular pentagon...

### 2 . Perimeter of a right triangle is equal to the sum of the diameter

ABC is a right angled triangle right angled at B.

Hence AC is diameter of the circumcircle as angle in a semi circle is a rig...

### 3 . ABC is right angle with angel B=90".

Since BC is diameter of the circle, ?BDC = 90° [Since angle in a semi circle is a right angle]

Consider ??s ABC and ADB...

### 4 . form an external point P tangent PA and PB are drawn to a circle .If CD is the tangent to the circle at a point E and PA=15cm,find the perimeter of the triangle PCD.

Sol :

Perimeter of ?PCD = 2 x AP.

=2 x 15 = 30 cm.

### 5 . Prove that ΔPRS ~ΔPTQ. If PQ = 4cm ,PT=3cm , ST= 5cm.

Sol:

PQ = 4 cm, PT = 3 cm , ST = 5 cm

(i)

PQ x PR = PT x PS

4 x PR = 3 x 8

PR = 6 cm

QR = PR...