Thales' Theorem

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: 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: 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: 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: 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: 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 $\angle ACD = 180° - 90° - 40° = 50°$.