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Statement:
If a line touches a circle and from the point of contact a chord is drawn, the angles between the tangent and the chord are respectively equal to the angles in the corresponding alternate segments.
Explanation:

In the figure, the chord $$PQ$$ divides the circle into two segments. Then, the tangent $$AB$$ is drawn such that it touches the circle at $$P$$.
Thus, the angle in the alternate segment for $$\angle QPB$$ is $$\angle QSP$$ and that for $$\angle QPA$$ is $$\angle PTQ$$ are equal.

Then, $$\angle QPB$$ $$=$$ $$\angle QSP$$ and $$\angle QPA$$ $$=$$ $$\angle PTQ$$.
Proof for the theorem:
Given:

Let $$O$$ be the centre of the circle.

The tangent $$AB$$ touches the circle at $$P$$, and $$PQ$$ is a chord.

Let $$S$$ and $$T$$ be the two points on the circle on the opposite sides of chord $$PQ$$.

To prove:

(i) $$\angle QPB$$ $$=$$ $$\angle QSP$$ and

(ii) $$\angle QPA$$ $$=$$ $$\angle PTQ$$

Construction:

Draw the diameter $$POR$$ and draw $$QR$$, $$QS$$ and $$PS$$.

Proof:

The diameter $$RP$$ is perpendicular to the tangent $$AB$$.

So, $$\angle RPB$$ $$=$$ $$90^{\circ}$$.

$$\Rightarrow$$ $$\angle RPQ$$ $$+$$ $$\angle QPB$$ $$=$$ $$90^{\circ}$$ …… $$(1)$$

Consider the triangle $$RPQ$$.

We know that:
Angle in a semicircle is $$90^{\circ}$$.
By the theorem, we have $$\angle RQP$$ $$=$$ $$90^{\circ}$$.  …… $$(2)$$

Thus, $$RPQ$$ is a right-angled triangle.

The sum of the two acute angles in the right-angled triangle, is $$90^{\circ}$$.

$$\Rightarrow$$ $$\angle QRP$$ $$+$$ $$\angle RPQ$$ $$=$$ $$90^{\circ}$$  …… $$(3)$$

From equations $$(1)$$ and $$(3)$$, we have:

$$\angle RPQ$$ $$+$$ $$\angle QPB$$ $$=$$ $$\angle QRP$$ $$+$$ $$\angle RPQ$$

$$\Rightarrow$$ $$\angle QPB$$ $$=$$ $$\angle QRP$$  …… $$(4)$$

We know that:
Angles in the same segment are equal.
By the theorem, we have $$\angle QRP$$ $$=$$ $$\angle PSQ$$. …… $$(5)$$

From equations $$(4)$$ and $$(5)$$, we have:

$$\angle QPB$$ $$=$$ $$\angle PSQ$$. …… $$(6)$$

Therefore, statement (i) is proved.

It is observed that $$\angle QPA$$ and $$\angle QPB$$ are linear angles.

We know that:
The linear pair of angle are always supplementary.
$$\Rightarrow$$ $$\angle QPA$$ $$+$$ $$\angle QPB$$ $$=$$ $$180^{\circ}$$ …… $$(7)$$

Consider the cyclic quadrilateral $$PSQT$$.

By the theorem, we have:
The sum of opposite angles of a cyclic quadrilateral is $$180^{\circ}$$.
$$\angle PSQ$$ $$+$$ $$\angle PTQ$$ $$=$$ $$180^{\circ}$$. …… $$(8)$$

From equations $$(7)$$ and $$(8)$$, we have:

$$\angle QPA$$ $$+$$ $$\angle QPB$$ $$=$$  $$\angle PSQ$$ $$+$$ $$\angle PTQ$$

Substitute equation $$(6)$$ in the above equation.

$$\Rightarrow$$ $$\angle QPA$$ $$+$$ $$\angle QPB$$ $$=$$  $$\angle QPB$$ $$+$$ $$\angle PTQ$$

$$\Rightarrow$$ $$\angle QPA$$ $$=$$ $$\angle PTQ$$

Therefore, statement (ii) is proved.
Example:

In the figure, $$AB$$ is the tangent, and $$CF$$ is the chord of the given circle. Next, find the value of $$x$$.

Solution:

Given, $$\angle BCF$$ $$=$$ $$57^{\circ}$$.

By the Alternate Segment Theorem, $$\angle BCF$$ $$=$$ $$\angle CDE$$.

Thus, $$x$$ $$=$$ $$57^{\circ}$$.

Therefore, the value of $$x$$ $$=$$ $$57^{\circ}$$.
Important!
This theorem is also known as Tangent - chord theorem.