In mathematics ,theorems are the foundational concepts that every student has to deal with before tackling any question in geometry. Without having a strong hold on theorems. it's quite difficult to understand or draw inferences from any geometrical shape.
In this article,we are going to describe an important theorem that combines the knowledge of circles, circle theorems, and triangles
So let's get started with the statement of alternate segment theorem along with its proof.
Alternate segment theorem
In any Circle, "The angle between a tangent and a chord of a circle through the point of contact is exactly equal to the angle subtended by the same chord in its alternate segment."
Taking a look at the figure , In mathematical terms , the \( \angle BAC \) and the \( \angle DBC \) must be equal.
Alternate segment theorem Proof
To begin with the proof of this theorem you must have the understanding of two basic theorems that are related to circles. The theorems are as follows:
Theorem-1 "Any line passing through the centre of the circle is always perpendicular to the tangent at the point of contact."
Theorem-2 "Angle subtended by an arc at the centre is always double of the measure of the angle subtended by the same arc at any other point of the remaining part of the circle."
If you want to get into the detailed explanation for these two theorems you can refer to our article Circle theorems
Proof : To prove \( \angle BAC = \angle DBC \)
Applying theorem 2 on the above diagram we can infer that
\(\angle BOC = 2 \angle BAC \)........Eq(1)
Applying theorem 1 on the above diagram we can write
\(\angle OBD = 90° \).........Eq(2)
Now consider triangle BOC inside the circle.
OB =OC (Both radii)
Thus \(\Delta BOC \) is an isosceles triangle.
\(\therefore \) \( \angle OBC = \angle OCB \)
\(\angle BOC + \angle OCB + \angle OBC\) =180°
\(\angle BOC + 2 \angle OBC \) =180°
From equation (1)
\(2\angle BAC + 2\angle OBC\) =180°
2\(\angle BAC + \angle OBC\) =180°
\(\angle BAC + \angle OBC\) = 90°
\( \angle OBC\) = 90°- \(\angle BAC \) .....Eq(3)
From equation (2) we can write.
\( \angle OBD = 90°\)
\( \angle OBC + \angle DBC = 90°\)
\( \angle DBC = 90° - \angle OBC \)
From equation (3) we can put the value of angle OBC
\( \angle DBC = 90° - ( 90°- \angle BAC ) \)
\( \angle DBC = \angle BAC \) [proved]
Alternate segment theorem Solved Examples
Example -1 : In the given circle, R is a point on the circumference. The line segment passing through R is a tangent to the circle. Given that\( \angle QPR = 35^\circ \), find the measure of \( \angle QRS \).
Ans : On applying alternate segment theorem we know that
\( \angle QPR = \angle QRS \)
\( \therefore \) The value of \( \angle QRS = 35^\circ \)
Example-2 : In the given circle, the line passing through T is a tangent to the circle. Given that the angle between the tangent and the chord TN is \( 20^\circ \) and that the lengths of chords MT and MN are equal, find the measure of \( \angle MNT \) .
Ans: By the Alternate Segment Theorem,
\( \angle TMN = \angle NTP = 20^\circ \)
We are given that MT = MN. This makes \( \triangle MTN \) an isosceles triangle,
Therefore, \( \angle MNT = \angle MTN \)
For \( \triangle MTN \)
The sum of angles in any triangle is \( 180^\circ \).
\( \angle TMN + \angle MNT + \angle MTN = 180^\circ \)
\( 20^\circ + 2(\angle MNT) = 180^\circ \)
\( 2(\angle MNT) = 160^\circ \)
\( \angle MNT = 80^\circ \)
Example- 3 : In the given circle with center O, the line passing through G is a tangent to the circle. Given that\( \angle OGF = 30^\circ \) where O is the center of the circle and G is the point of tangency, find the measure of \( \angle FGH \).
Ans : A radius drawn to the point of tangency is always perpendicular to the tangent line.
Therefore,\( \angle OGH = 90^\circ \)
\( \angle OGH = \angle OGF + \angle FGH = 90^\circ \)
We are given \( \angle OGF = 30^\circ \)
\( 30^\circ + \angle FGH = 90^\circ \)
\( \angle FGH = 90^\circ - 30^\circ \)
\( \angle FGH = 90^\circ - 30^\circ \)
\( \angle FGH = 60^\circ \)
