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Use the hyperboloid model with Minkowski inner productPut the vertex with angle $\theta$ at $v=(1,0,0)$, and place the endpoints of its adjacent sides of lengths $a$ and $b$ atThe geodesic through $P$ perpendicular to $vP$ has spacelike unit normalwhile the geodesic through $Q$ perpendicular to $vQ$ has spacelike unit normal

These two geodesics form the remaining two sides of the quadrilateral. Their angle equals the angle between their normals in the tangent plane at their intersection. The third right angle therefore givesDividing by $\cosh a\cosh b$ proves the Lambert quadrilateral identity