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# Alternate segment theorem

#### Summary

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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.

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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.
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### 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]

### 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...