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Hint: In this question, we are asked to state and prove the Thales theorem. Therefore, we should state the theorem and then make an appropriate diagram and prove the theorem.
Complete step-by-step answer:
Thales theorem can be stated as
If P, Q and R are three distinct points on a circle, and if the line segment joining P and R forms a diameter of the circle (i.e. it passes through the center of the circle), then $\angle PQR={{90}^{\circ }}$.
To prove this theorem, we can construct the figure as shown below in which P, Q, R are three points on the circle with center O and PR is the diameter.
We need to show that $\angle PQR={{90}^{\circ }}$.
For this, we use the following theorem
The angles facing equal sides in a triangle have the same value………………………… (1.1)
In the figure, we find that as P, Q and R are points on a circle
\[\begin{align}
& \text{In }\Delta OPQ, \\
& OP=OQ=\text{radius of the circle} \\
& \Rightarrow \angle \text{OPQ=}\angle \text{PQO (from equation 1}\text{.1)}...........\text{(1}\text{.2)} \\
\end{align}\]
And
\[\begin{align}
& \text{In }\Delta ORQ, \\
& OR=OQ=\text{radius of the circle} \\
& \Rightarrow \angle \text{ORQ=}\angle R\text{QO (from equation 1}\text{.1)}............\text{(1}\text{.3)} \\
\end{align}\]
Now, from the figure, we see that
$\begin{align}
& \angle PQR=\angle PQO+\angle RQO \\
& \Rightarrow \angle PQR=\angle OPQ+\angle ORQ\text{ (from equations 1}\text{.2 and 1}\text{.3)}................(1.4) \\
\end{align}$
Now, we can use the angle sum property of a triangle which states that
The sum of the three interior angles of a triangle is equal to ${{180}^{\circ }}$…………………..(1.5)
Now, using equation (1.5) in $\Delta PQR$,
$\begin{align}
& \angle PQR+\angle OPQ+\angle ORQ={{180}^{\circ }} \\
& \Rightarrow \angle PQR+\angle PQR={{180}^{\circ }}\Rightarrow \angle PQR=\dfrac{{{180}^{\circ }}}{2}={{90}^{\circ }} \\
\end{align}$
Thus, we have proved that $\angle PQR={{90}^{\circ }}$ as stated in the Thales theorem and thus have proved it.
Note: We should note that the proof would not be valid if PR is not the diameter, as then PR would not pass through the center of the circle O and thus equations (1.2) and (1.3) will not be valid.
Complete step-by-step answer:
Thales theorem can be stated as
If P, Q and R are three distinct points on a circle, and if the line segment joining P and R forms a diameter of the circle (i.e. it passes through the center of the circle), then $\angle PQR={{90}^{\circ }}$.
To prove this theorem, we can construct the figure as shown below in which P, Q, R are three points on the circle with center O and PR is the diameter.
We need to show that $\angle PQR={{90}^{\circ }}$.
For this, we use the following theorem
The angles facing equal sides in a triangle have the same value………………………… (1.1)
In the figure, we find that as P, Q and R are points on a circle
\[\begin{align}
& \text{In }\Delta OPQ, \\
& OP=OQ=\text{radius of the circle} \\
& \Rightarrow \angle \text{OPQ=}\angle \text{PQO (from equation 1}\text{.1)}...........\text{(1}\text{.2)} \\
\end{align}\]
And
\[\begin{align}
& \text{In }\Delta ORQ, \\
& OR=OQ=\text{radius of the circle} \\
& \Rightarrow \angle \text{ORQ=}\angle R\text{QO (from equation 1}\text{.1)}............\text{(1}\text{.3)} \\
\end{align}\]
Now, from the figure, we see that
$\begin{align}
& \angle PQR=\angle PQO+\angle RQO \\
& \Rightarrow \angle PQR=\angle OPQ+\angle ORQ\text{ (from equations 1}\text{.2 and 1}\text{.3)}................(1.4) \\
\end{align}$
Now, we can use the angle sum property of a triangle which states that
The sum of the three interior angles of a triangle is equal to ${{180}^{\circ }}$…………………..(1.5)
Now, using equation (1.5) in $\Delta PQR$,
$\begin{align}
& \angle PQR+\angle OPQ+\angle ORQ={{180}^{\circ }} \\
& \Rightarrow \angle PQR+\angle PQR={{180}^{\circ }}\Rightarrow \angle PQR=\dfrac{{{180}^{\circ }}}{2}={{90}^{\circ }} \\
\end{align}$
Thus, we have proved that $\angle PQR={{90}^{\circ }}$ as stated in the Thales theorem and thus have proved it.
Note: We should note that the proof would not be valid if PR is not the diameter, as then PR would not pass through the center of the circle O and thus equations (1.2) and (1.3) will not be valid.
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