Proof concerning circles

Question

How do I prove that the diameter of a circle subtends a right angle at a circumference? Thank you in advance! I haven't got the slightest idea.

Answer 1

The standard way is as follows. Add a segment from the intersection with the circumference to the center. You get two isosceles triangles. Use the following facts to finish the proof:

1. In an isoceles triangle, the two base angles are the same.

2. The angle around the center at the diameter is twice a right angle.

3. The sum of angles in a triangle is twice a right angle.

Answer 2

One among Euclid's earlier theorems states that angle subtended by a chord at any point of opposite arc is half that at center. Taking the special case of the biggest chord which is nothing but the diameter of the circle, it subtends 2 right angle at center and one right angle at any point on the semi-circle.

Answer 3

Using vectors, take a unit circle centred on the origin, take the two radii vectors that make up the diameter as $\pm \vec i$ and the vector from the centre to some point on the circumference as $\vec r$. The two vectors from the edges are $r\pm i$ and their dot product is $r^2-i^2$ which is zero as the magnitudes of both $i$ and $r$ are 1. Since the dot product is zero these two vectors are perpendicular.