0

Here is the answer:

Let l be the h-symmetry of OA (O is center of absolute).

In respect to l, A goes to O and B goes to some point B'.

Now we construct a circle c(0,B')

Then we apply l again and c(O,OB') goes to c(A,AB)

We now find h-symmetry of OB and similarly we get circle c(B,BA)

Now point C is intersection of c(A,AB) and c(B,BA)

My question is why is the angle between c(A,AB) and c(B,BA) 90 degrees?

In euclidean geometry it would be impossible since both circles contain the center of the other one.

Also inscribed angle of semicircle is less than 90 degrees in hyperbolic geometry, shouldn't angle ACB be even less than that?

JurgenKlop
  • 183

1 Answers1

1

To summarize the construction you described on the hyperbolic plane:

  • you construct a circle around $A$ through $B$
  • you construct a circle around $B$ through $A$
  • you choose an intersection point $C$ of those two circles.

The resulting triangle $ABC$ is equilateral by construction, so all its angles are equal. Since there are no triangles with three right angles in the hyperbolic plane, $ABC$ is not a right-angled triangle.

However, you can modify this construction to obtain an appropriate triangle: Simply let $D$ be the midpoint between $A$ and $C$, and have a look at the triangle $ABD$ instead. The angle $ADB$ is a right angle, because the angles $ADB$ and $BDC$ sum to $\pi$ and are equal by symmetry.

Magma
  • 6,270