$ABCD$ 是一个平行四边形，且三角形 $CDE$ 是等腰三角形。
$E$ 和 $F$ 分别是 $AD$ 和 $BC$ 上的点。
$AB$、$EF$ 和 $CD$ 是平行线。
角 $BAE = 3x^\circ$ 且角 $CEF = 1.5x^\circ$。 求 $x$ 的值。请清楚地写出理由。 正 $n$ 边形的一个内角与正 $(n + 2)$ 边形的一个内角的比是 15 : 16。求 $n$ 的值。 第 9(b) 题的解题空间续 答 $n =$ ... $ABCD$ 是一块花园绿地。 计划在这座花园中建造一个游乐场 $P$，该游乐场到边 $AB$ 和 $BC$ 的距离相等，且到点 $A$ 和点 $D$ 的距离也相等。 请用尺规作图画出 $P$ 的位置。 已知该慈善活动赚取的所有利润都将捐赠给志愿福利组织，她的捐款是否真的超过了她成本价的 900%？
请给出理由并清楚地写出所有的计算过程。

1. Find $\angle CDE$ using interior angles between parallel lines.

Since $ABCD$ is a parallelogram, $AB \parallel CD$.
$\angle CDE = 180^\circ - \angle BAE$ (interior angles, $AB \parallel CD$)
$\angle CDE = 180^\circ - 3x^\circ$

2. Express $\angle CED$ using the isosceles triangle property.

Triangle $CDE$ is isosceles with $CD = CE$.
$\angle CED = \angle CDE = 180^\circ - 3x^\circ$ (base angles of isosceles triangle)

3. Use parallel lines $EF$ and $CD$ to find $\angle FED$ and form an equation to solve for $x$.

Since $EF \parallel CD$,
$\angle FED = 180^\circ - \angle CDE$ (interior angles, $EF \parallel CD$)
$\angle FED = 180^\circ - (180^\circ - 3x^\circ) = 3x^\circ$

From the diagram, $\angle FED = \angle CEF + \angle CED$.
$3x^\circ = 1.5x^\circ + (180^\circ - 3x^\circ)$
$3x = 180 - 1.5x$
$4.5x = 180$
$x = 40$

1. Express the interior angles of both polygons in terms of $n$.

Interior angle of a regular $n$-sided polygon = $\frac{(n-2) \times 180^\circ}{n}$
Interior angle of a regular $(n+2)$-sided polygon = $\frac{(n+2-2) \times 180^\circ}{n+2} = \frac{n \times 180^\circ}{n+2}$

2. Set up the equation based on the given ratio.

$\frac{\frac{(n-2) \times 180}{n}}{\frac{n \times 180}{n+2}} = \frac{15}{16}$
$\frac{n-2}{n} \times \frac{n+2}{n} = \frac{15}{16}$
$\frac{n^2 - 4}{n^2} = \frac{15}{16}$

3. Solve the equation for $n$.

$16(n^2 - 4) = 15n^2$
$16n^2 - 64 = 15n^2$
$n^2 = 64$
Since $n > 2$, $n = 8$.

1. Construct the locus of points equidistant from sides $AB$ and $BC$.

Using a pair of compasses and a straightedge, construct the angle bisector of $\angle ABC$. Draw a line from $B$ through the intersection of the arcs.

2. Construct the locus of points equidistant from points $A$ and $D$.

Using a pair of compasses and a straightedge, construct the perpendicular bisector of the line segment $AD$. Draw a line through the intersection of the arcs above and below $AD$.

3. Locate and mark point $P$.

Mark the point of intersection of the angle bisector of $\angle ABC$ and the perpendicular bisector of $AD$ as $P$.