3. In a right-angled trapezoid, the length of the midline is 43. The shorter diagonal of this trapezoid is also the angle bisector of the obtuse angle of the trapezoid, and its length is 60. Determine the lengths of all sides of this trapezoid.
3. Let the right angles of the trapezoid be at $A$ and $D$, $AB$ the larger base, and $CD$ the smaller. According to the problem, we have $\measuredangle ACB = \measuredangle ACD = \varphi$, and from this it follows that $\measuredangle CAB = \measuredangle ACD = \varphi$, as alternate interior angles. Thus, $\triangle ABC$ is isosceles, i.e., $AB = BC = c$. Let $O$ be the intersection of the midline of the trapezoid with the diagonal $AC$; hence, $O$ is the midpoint of $AC$. Since $\triangle ADC$ is a right triangle, we have $OD = OA = OC = \frac{AC}{2} = 30$. Since $\triangle OCD$ is also isosceles with the base angle $\varphi$, we have $\triangle ABC \sim \triangle DOC$. If we denote $CD = x$, we get the proportion $\frac{60}{c} = \frac{x}{30}$, from which it follows that $cx = 1800$. Further, we have $c + x = AB + CD = 2 \cdot 43 = 86$, so $c = 86 - x$, and substituting this into the previous equation, we get $(86 - x)x = 1800$, i.e., $x^2 - 86x + 1800 = 0$. Solving this equation, we find $x_{1/2} = \frac{86 \pm \sqrt{86^2 - 4 \cdot 1800}}{2} = \frac{86 \pm \sqrt{196}}{2} = \frac{86 \pm 14}{2}$, i.e., $x = 36$ and $c = 86 - x = 50$ (we discard the other solution: $x = 50$ and $c = 36$, because $c > x$ must hold). Thus, we have $AB = BC = 50$, $CD = x = 36$, and using the Pythagorean theorem, we calculate $AD = \sqrt{AC^2 - CD^2} = \sqrt{60^2 - 36^2} = \sqrt{3600 - 1296} = \sqrt{2304} = 48$.

![](https://cdn.mathpix.com/cropped/2024_06_04_841f52b4ffb037ad799bg-34.jpg?height=497&width=476&top_left_y=679&top_left_x=1050)

Op 2018 2B 3 of this into the previous equation, we get $(86 - x)x = 1800$, i.e., $x^2 - 86x + 1800 = 0$. Solving this equation, we find $x_{1/2} = \frac{86 \pm \sqrt{86^2 - 4 \cdot 1800}}{2} = \frac{86 \pm \sqrt{196}}{2} = \frac{86 \pm 14}{2}$, i.e., $x = 36$ and $c = 86 - x = 50$ (we discard the other solution: $x = 50$ and $c = 36$, because $c > x$ must hold). Thus, we have $AB = BC = 50$, $CD = x = 36$, and using the Pythagorean theorem, we calculate $AD = \sqrt{AC^2 - CD^2} = \sqrt{60^2 - 36^2} = \sqrt{3600 - 1296} = \sqrt{2304} = 48$.