In geometry, Thales' theorem is stated by Greek mathematician Thales of Ionia. He is dubbed as father of geometry for his great contribution in geometry theorems.
He attempts to establish and identify various theorems in geometry. He also credited as the first to explicitly detail a logical proof of a geometric result.
Nowadays this result is stated as, If an angle is drawn from the diameter of a circle to a point on its circumference, then the formed angle is always a right angle.
In this section we will discuss Thales theorem in detail with its proof and some solved examples.

## Statement

In geometry, Thales' theorem states that "whenever an angle is drawn from the diameter of the circle to the point on its circumference, then angle formed is sure to be a perfect right angle."Â

In other words, if A,B and C are the points on a circle and AC is the diameter, then $\angle$ ABC is of 90 degree i.e., a right angle.

## Proof

While keeping the statement of theorem in mind, lets prove the theorem:
Suppose that $\angle$A = $\alpha$ and $\angle$C = $\beta$. Then $\angle$ B = $\alpha$ + $\beta$

Thales was at point of his understanding in geometry to believe two things:
1) The three interior angles of any triangle sum to 180 degree.
2) The two base angles of an isosceles triangle are congruent.
As we can see in above figure:
$\angle$ A + $\angle$ B + $\angle$ C = 180$^0$
$\alpha$ + ($\alpha$ + $\beta$) + $\beta$ = 180$^0$

Thus,
2$\alpha$ + 2$\beta$ = 180$^0$
Giving,
$\alpha$ + $\beta$ = 90$^0$
$\angle$B = 90$^0$
we have a right angle.

## Thales Theorem Triangle

If we talk about Thales theorem triangle, we will familiar with an important truth about equiangular triangle. Thales stated this theorem as ratio of any two corresponding sides of an equiangular triangle is always same irrespective of their sizes. And this is known as BPT (Basic Proportionality Theorem).

If we draw a line parallel to any one arm of a triangle which passes through the other two arms, then other arms are divided in the same ratio.

If DE || BC, then

$\frac{AE}{EC}$$\frac{AD}{DB}$

## Examples

Example 1: From the below figure, find the value of x.

Solution:
Given that $\angle$ C = 30$^0$
From the Thales theorem, if an angle is drawn from the diameter of the circle to the point on its circumference, then angle formed is a right angle.

=> $\angle$ A = 90$^0$   ......(1)

Again we know, sum of angles of a triangle is equal to 180$^0$

=> $\angle$A + $\angle$B + $\angle$C = 180$^0$  .......(2)
Using (1), we have
90 + x + 30 = 180
x = 180 - 120 = 60
or x = 60
The value of x is 60$^0$

Example 2:

In figure, if PQ||CB and PR||CD.

Prove that $\frac{AQ}{QB}$=$\frac{AR}{RD}$

Solution:

In triangle ABC it is given that PQ||BC.

Therefore  $\frac{AQ}{QB}$=$\frac{AP}{PC}$       { by BPT}.....(1)

In triangle ACD it is given that RP||CD.

Therefore $\frac{AP}{PC}$=$\frac{AR}{RD}$            {by BPT}....(2)

From (1) and (2) we can say that,

$\frac{AQ}{QB}$=$\frac{AR}{RD}$

Hence Proved.