[color=#ff0000]In the hyperbolic plane[/color][color=#333333] g[/color]iven circle [i]s [/i](with center [i]P [/i]and a circumpoint), and line [i]a[/i]. Construct the lines perpendicular to [i]a[/i] and tangential to [i]s[/i].

In Euclidean geometry this is not a real challenge — there is only one line through [i]P[/i] parallel to [i]a[/i]. But in our case the problem is much more complex, so it is worth applying [url=https://en.wikipedia.org/wiki/George_P%C3%B3lya]George Pólya[/url]'s (1887-1985) [url=https://en.wikipedia.org/wiki/How_to_Solve_It]method[/url], namely, we assume what is required to be done as already done. Draw tangent [i]e[sub]a [/sub][/i]to the given circle [i]s[/i] through an arbitrary point [i]E[sub]A[/sub][/i], consider an arbitrary point [i]A [/i]on it, and the line through [i]A[/i] and perpendicular to [i]e[sub]a[/sub][/i]. This line will be used as an input in another problem later. Let [i]T[/i] be the perpendicular projection of center [i]P[/i] of circle [i]s[/i] on line [i]a[/i]. [br][br]By recalling Bolyai's construction (Fig. 4) we recognize a Lambert quadrilateral in the construction, with three right angles in quadrilateral [i]PE[sub]A[/sub]AT[/i].[br]

In the present case, firstly we construct the perpendicular projection [i]T[/i] of center [i]P[/i] on [i]a[/i]. Next, the intersection point [i]A[/i][sub]0[/sub] of circle [i]s[/i] and ray [[i]P,V[/i][sub]1[/sub]) asymptotically parallel to [i]a [/i]are constructed (Fig. 6b). The orthogonal projection of point [i]A[/i][sub]0 [/sub]to line ([i]PT[/i]) yields point [i]T[/i][sub]0[/sub]. At this stage we have two options to continue:[br][br][list][*]Point [i]E[sub]A[/sub][/i] is obtained by intersecting circle [i]s[/i] and ray [[i]PV[/i][sub]0[/sub]) — here point [i]V[/i][sub]0[/sub] is a point of line ([i]A[/i][sub]0[/sub][i]T[/i][sub]0[/sub]) in infinity. In this case we must show that line [i]e[sub]A[/sub] [/i]through [i]E[sub]A [/sub][/i]and tangential to [i]s[/i] is perpendicular to [i]a[/i]. [br][br][/*][*]Let [i]A[/i] be the mirror image of point [i]A[/i][sub]0 [/sub]with respect to the perpendicular bisector of segment [i]T[/i][sub]0[/sub][i]T[/i], and line [i]e[sub]A[/sub][/i] the perpendicular line through [i]A[/i] to [i]a.[/i] In this case we must show that [i]e[sub]A[/sub][/i] is tangential to circle [i]s[/i], that is, the perpendicular projection of [i]P [/i]on [i]e[sub]A [/sub][/i]goes through the circle [i]s[/i].[/*][/list]