如圖所示，設\(ABCD\)為等腰梯形，其中\(\overline{BC}//\overline{AD}\)且\(\overline{AB}=\overline{CD}\)，點\(X\)與點\(Y\)在對角線\(\overline{AC}\)上，其中點\(X\)位於點\(A\)與點\(Y\)之間。若\(\angle AXD=\angle BYC=90^{\circ}\)，\(\overline{AX}=3\)，\(\overline{XY}=1\)，\(\overline{YC}=2\)，則\(ABCD\)的面積為多少？
[解答]
BD=AC=6,BY+XD=√(BD^2-XY^2)=√35
所求=1/2×AC×√35=3√35
113.3.17補充
Let \(ABCD\) be an isosceles trapezoid with \(\overline{BC} \parallel \overline{AD}\) and \(AB=CD\). Points \(X\) and \(Y\) lie on diagonal \(\overline{AC}\) with \(X\) between \(A\) and \(Y\), as shown in the figure. Suppose \(\angle AXD = \angle BYC = 90^\circ\), \(AX = 3\), \(XY = 1\), and \(YC = 2\). What is the area of \(ABCD\)?
(2021Fall AMC12A，
https://artofproblemsolving.com/ ... Problems#Problem_21)
(2021秋季AMC12A，
https://math.pro/db/thread-3591-1-2.html)
