Thales' Theorem

By Thales' Theorem, if AB is a diameter and C is any other point on the circle, then $\angle ACB=90^{\circ }$.

One way to see why: the angle at the centre, $\angle AOB$, is a straight line, so it is $180^{\circ }$. The angle at the circumference standing on the same arc is half of it, $180^{\circ }÷2=90^{\circ }$.

This is true wherever C sits on the circle, so no measuring and no assumptions about the triangle's shape are needed.

By Thales' Theorem, if AB is a diameter and C is any other point on the circle, then $\angle ACB=90^{\circ }$.

The angle at the centre, $\angle AOB$, is a straight line and therefore $180^{\circ }$; the angle at the circumference on the same arc is half of it, so $\angle ACB=180^{\circ }÷2=90^{\circ }$.

It makes no difference that C is on the lower arc here: the result holds wherever C sits on the circle, and needs no measuring.

Since AB is the diameter, Thales' Theorem gives $\angle ACB=90^{\circ }$.

The angles in triangle ABC add up to $180^{\circ }$, so $\angle ABC=180^{\circ }-90^{\circ }-65^{\circ }=25^{\circ }$.

Two traps: the $90^{\circ }$ is only the first step (the question asks for the angle at B, not at C), and nothing says the triangle is isosceles. The angles at A and B are $65^{\circ }$ and $25^{\circ }$, not equal.

Since AB is the diameter, Thales' Theorem gives $\angle ACB=90^{\circ }$.

The angles in triangle ABC add up to $180^{\circ }$, so $\angle CAB=180^{\circ }-90^{\circ }-35^{\circ }=55^{\circ }$.

Check: $55^{\circ }+35^{\circ }+90^{\circ }=180^{\circ }$. Notice the angles at A and B are not equal: nothing here says the triangle is isosceles.

Thales' Theorem says that when AB is the diameter and C is on the circle, $\angle ACB=90^{\circ }$, whatever the other angles are.

So the $50^{\circ }$ at A is not needed here. It would only be needed for the third angle: $\angle ABC=180^{\circ }-90^{\circ }-50^{\circ }=40^{\circ }$.

The answer is $\angle ACB=90^{\circ }$. The middle letter, C, tells you the angle is at C, not at A.

By Thales' Theorem, AB is a diameter so $\angle ACB=90^{\circ }$.

In triangle ABC: $\angle CAB=180^{\circ }-90^{\circ }-50^{\circ }=40^{\circ }$.

CD is perpendicular to AB, so in triangle ACD we have $\angle ADC=90^{\circ }$ and $\angle CAD=40^{\circ }$, giving $\angle ACD=180^{\circ }-90^{\circ }-40^{\circ }=50^{\circ }$.

Note that CD does not cut the right angle at C in half: $\angle ACD=50^{\circ }$ while $\angle BCD=40^{\circ }$.