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Thales’ Theorem

Geometry and Measures Thales’ Theorem

Courses Overview

🎬 Video: Thales' Theorem Definition and Application

What is Thales' Theorem? (0:01)

Thales’s theorem states that if a triangle is formed using the diameter of a circle and any point on the semicircle, it will always be a right-angled triangle.

In the diagram, AB is the diameter of the circle, and C is a point on the semicircle. No matter where C is, the angle at C is always 90°.

How to Apply Thales' Theorem? (0:37)

Step 1: Use Thales’s Theorem
Since $AB$ is the diameter, and vertex $C$ is a point on the semicircle. Thus,

$$ \angle C = 90^\circ $$

Step 2: Find the Remaining Angle
The sum of all angles in a triangle is:

$$ \angle A + \angle B + \angle C = 180^\circ $$

Since we know $ \angle C = 90^\circ $ and $ \angle A = 35^\circ $, we can find $ \angle B $:

$$ \angle B = 180^\circ – 90^\circ – 35^\circ = 55^\circ $$

📂 Flashcards: Thales' Theorem Examples with Triangles

Explanation of Thales' Theorem with a diagram showing a right-angled triangle formed by a diameter and any point on a semicircle. — 1) What is Thales' Theorem?

Diagram for Thales' Theorem with triangle inscribed in a semicircle, showing a right angle at point C and interior angles A = 35° and B = 55°. — 2) How to Apply Thales' Theorem

Semicircle with diameter AB, triangle ACB showing angle ACB as 90° using Thales's theorem, with marked angles 40°, 50°, and 90°. — 3) Thales Theorem: Exercise

🍪 Quiz: Test Your Skills with Thales' Theorem

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Thales' Theorem

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Q: In the following diagram, AB is the diameter of a semicircle, and C is a point on the circle. What is $\angle ACB$?

$60^\circ$

$45^\circ$

$90^\circ$

$120^\circ$

Feedback: Not quite! By Thales's Theorem, we know that when we have a circle with a diameter AB and any point C on the semicircle, the angle $\angle ACB$ will always be a right angle, regardless of the location of point C on the semicircle.

Feedback: Well done! By Thales's Theorem, we know that when we have a circle with a diameter AB and any point C on the semicircle, the angle $\angle ACB$ will always be a right angle, regardless of the location of point C on the semicircle.

2 / 6

Q: In the following diagram, AB is the diameter of a circle, and C is a point on the circle. What is $\angle ACB$?

$120^\circ$

$45^\circ$

$90^\circ$

$60^\circ$

Feedback: Not quite! By Thales's Theorem, we know that when we have a circle with a diameter AB and any point C on the semicircle, the angle $\angle ACB$ will always be a right angle, regardless of the location of point C on the semicircle.

Feedback: Well done! By Thales's Theorem, we know that when we have a circle with a diameter AB and any point C on the semicircle, the angle $\angle ACB$ will always be a right angle, regardless of the location of point C on the semicircle.

3 / 6

Q: In the following diagram, AB is the diameter of a semicircle, and C is a point on the circle. If $\angle CAB = 65°$, find $\angle ABC$

$65^\circ$

$85^\circ$

$90^\circ$

$25^\circ$

Feedback: Not quite! Since AB is the diameter of the semicircle, Thales’s Theorem tells us that $\angle ACB = 90°$. We are given that $\angle CAB = 65°$. In any triangle, the sum of the angles is 180°. Therefore, to find $\angle ABC$, we use: $\angle ABC = 180° - 90° - 65° = 25°$.

Feedback: Well done! Since AB is the diameter of the semicircle, Thales’s Theorem tells us that $\angle ACB = 90°$. We are given that $\angle CAB = 65°$. In any triangle, the sum of the angles is 180°. Therefore, to find $\angle ABC$, we use: $\angle ABC = 180° - 90° - 65° = 25°$.

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Q: In the following diagram, AB is the diameter of a semicircle, and C is a point on the circle. If $\angle ABC = 35°$, find $\angle CAB$.

$90^\circ$

$55^\circ$

$65^\circ$

$35^\circ$

Feedback: Not quite! Thales’s Theorem tells us that $\angle ACB = 90°$. We are given that $\angle ABC = 35°$, and we know that the sum of the angles in any triangle is 180°. To find $\angle CAB$, we use: $\angle CAB = 180° - 90° - 35° = 55°$.

Feedback: Well done! Thales’s Theorem tells us that $\angle ACB = 90°$. We are given that $\angle ABC = 35°$, and we know that the sum of the angles in any triangle is 180°. To find $\angle CAB$, we use: $\angle CAB = 180° - 90° - 35° = 55°$.

5 / 6

Q: In the following diagram, AB is the diameter of a semicircle, and C is a point on the circle. If $\angle CAB = 50°$, find $\angle ACB$.

$90^\circ$

$40^\circ$

$50^\circ$

$60^\circ$

Feedback: Not quite! Thales's Theorem states that when AB is the diameter of a semicircle, $\angle ACB$ will always be 90°, no matter the other angles. Thus, $\angle ACB$ = 90°.

Feedback: Well done! Thales's Theorem states that when AB is the diameter of a semicircle, $\angle ACB$ will always be 90°, no matter the other angles. Thus, $\angle ACB$ = 90°.

6 / 6

Q: In the following diagram, AB is the diameter of a semicircle, and C is a point on the circle. If $\angle B = 50°$, find $\angle ACD$.

$60^\circ$

$40^\circ$

$90^\circ$

$50^\circ$

Feedback: Not quite! By Thales’s Theorem, since AB is the diameter of the semicircle, we know that $\angle ACB = 90°$. Then, in right-angled triangle ABC, we can calculate $\angle A = 180° - 90° - 50° = 40°$. Finally, in right-angled triangle ACD, we find $\angle ACD = 180° - 90° - 40° = 50°$.

Feedback: Well done! By Thales’s Theorem, since AB is the diameter of the semicircle, we know that $\angle ACB = 90°$. Then, in right-angled triangle ABC, we can calculate $\angle A = 180° - 90° - 50° = 40°$. Finally, in right-angled triangle ACD, we find $\angle ACD = 180° - 90° - 40° = 50°$.

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