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1. Construct two geodeters, l1 and l2, in Poincar´e’s upper half plane (PÖH) that do not intersect. Then draw a geodet m that intersects both l1 and l2 and where the alternate angles are different. As before, the alternate angles are on each side of m, but this time it is angles between tangents at the intersection points. You have thus shown that Euclid’s 29th theorem does not apply in HG. This is where he got use of his fifth axiom that does not apply in HG. Theorem 27, on the other hand, also applies in HG. It says that if the alternate angles are equal, then the two geodeters cannot intersect each other.

2. In this problem you need the equation of the circle. Given the surveyor, m, which passes through (- √ 1 2, √ 1 2) and (√ 1 2, √ 1 2) in PÖH and a point A = (0, 2). Find the geode (circular arc), n, through A that meets (tangents) m on the positive x-axis (i.e. the infinity in the model). The angle between the y-axis and the key to n at point A is called the parallel angle. For smaller angles, cut geodeters through the A geodesy m. Calculate the parallel angle.

3. Show that a circle c1 passing through point A and its inverted point B is orthogonal to circle c with the radius OP. See figure. The fact that two circles are orthogonal to each other means that the angle between the keys at the point of intersection is 90 degrees. We need this result in the next task.

Point B is inverse to point A in the circle C. Are the triangles OBP and OPA uniform? Are the angles OBP and APO equal? If the circle C1 is orthogonal to C, it means that the angle O1PO is right.



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