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 = $3x^\circ$，角 CEF = $1.5x^\circ$。 求 x 的值。清楚地陈述你的理由。 答案 x = ... 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。 答案 n = ... ABCD 是一块花园的土地。 计划在花园里建造一个游乐场 P，使其到边 AB 和 BC 的距离相等，且到点 A 和点 D 的距离相等。 通过作图，标出 P 的位置。

1. Find the base angle at D in triangle CDE using the parallelogram/parallel-line relationship.

Since $AB \parallel CD$ and $AE$ lies on the same straight line as $DE$, the angles $\angle BAE$ and $\angle CDE$ are co-interior angles.

$\angle CDE = 180^\circ - 3x^\circ$

2. Use the isosceles triangle property and parallel lines to identify the other two angles in triangle CDE.

In triangle $CDE$, $CD = CE$, so the base angles are equal:

$\angle CDE = \angle CED = 180^\circ - 3x^\circ$

Also, since $EF \parallel CD$,

$\angle ECD = \angle CEF = 1.5x^\circ$

corresponding/alternate angles are equal.

3. Apply the angle sum of triangle CDE and solve for x.

Using the angle sum of triangle $CDE$:

$(180 - 3x) + (180 - 3x) + 1.5x = 180$

$360 - 6x + 1.5x = 180$

$360 - 4.5x = 180$

$4.5x = 180$

$x = 40$

1. Write expressions for the interior angles of the two regular polygons.

Interior angle of a regular $m$-sided polygon is
$\frac{(m-2)180^\circ}{m}.$
For the regular $n$-sided polygon:
$\frac{(n-2)180}{n}.$
For the regular $(n+2)$-sided polygon:
$\frac{((n+2)-2)180}{n+2}=\frac{180n}{n+2}.$

2. Form an equation using the given ratio $15:16$.

$\frac{\frac{(n-2)180}{n}}{\frac{180n}{n+2}}=\frac{15}{16}$
Cancel $180$:
$\frac{(n-2)(n+2)}{n^2}=\frac{15}{16}.$

3. Solve the resulting quadratic equation and choose the valid polygon value.

$16(n-2)(n+2)=15n^2$
$16(n^2-4)=15n^2$
$16n^2-64=15n^2$
$n^2=64$
$n=8 \text{ or } -8.$
Since the number of sides of a polygon must be positive,
$n=8.$

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

Using a compass and ruler, construct the internal angle bisector of $\angle ABC$.

This line is the locus of points equidistant from the sides $AB$ and $BC$.

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

Using a compass and ruler, construct the perpendicular bisector of $AD$.

This line is the locus of points equidistant from $A$ and $D$.

3. Mark the required point $P$.

Mark the point of intersection of the two constructed lines and label it $P$.