1. **Stating the problem:** We are given points of a quadrilateral with vertices \(A(1,3), B(4,1), C(6,4), D(3,6)\) and vectors \(\vec{u} = \overrightarrow{AB} = (3, -2)\) and \(\vec{v} = \overrightarrow{DC} = (-3, 2)\). We want to verify properties of the figure, check orthogonality of segments, and find the midpoint \(L\) of the diagonals. 2. **Checking lengths of segments \(AB\) and \(DC\):** Length formula: \(\parallel \vec{w} \parallel = \sqrt{w_x^2 + w_y^2}\). Calculate \(\parallel AB \parallel = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\). Calculate \(\parallel DC \parallel = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}\). So, \(\parallel AB \parallel = \parallel DC \parallel = \sqrt{13}\), not \(\sqrt{5}\) as initially stated. 3. **Checking orthogonality of \(\vec{u}\) and \(\vec{v}\):** Dot product formula: \(\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y\). Calculate \(\vec{u} \cdot \vec{v} = 3 \times (-3) + (-2) \times 2 = -9 - 4 = -13 \neq 0\). Since the dot product is not zero, \(\vec{u}\) and \(\vec{v}\) are not orthogonal. 4. **Checking if triangle \(ADC\) is right angled:** Vectors \(\overrightarrow{AC} = (6-1, 4-3) = (5,1)\) and \(\overrightarrow{DC} = (-3, 2)\). Calculate \(\overrightarrow{AC} \cdot \overrightarrow{DC} = 5 \times (-3) + 1 \times 2 = -15 + 2 = -13 \neq 0\). So, \(\overrightarrow{AC}\) and \(\overrightarrow{DC}\) are not perpendicular, so triangle \(ADC\) is not right angled at \(D\). 5. **Finding midpoint \(L\) of diagonals \(AC\) and \(BD\):** Midpoint formula: \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\). Calculate \(L = \frac{A + C}{2} = \left(\frac{1 + 6}{2}, \frac{3 + 4}{2}\right) = (3.5, 3.5)\). Calculate \(L = \frac{B + D}{2} = \left(\frac{4 + 3}{2}, \frac{1 + 6}{2}\right) = (3.5, 3.5)\). Since both midpoints coincide, \(L\) is the intersection of diagonals and the center of the figure. 6. **Explanation about points \(T_1\) and \(T_3\):** They are midpoints of sides \(AB\) and \(DC\), which are parallel sides of the quadrilateral. **Final conclusion:** - Segments \(AB\) and \(DC\) have equal length \(\sqrt{13}\). - They are not orthogonal. - The diagonals intersect at \(L(3.5, 3.5)\), the midpoint of both diagonals.