请帮我解答这道题---**Extraction Content:**
**[定义]** 已知两个矩形，若其中一个矩形的四个顶点分别在另一个矩形的四条边上 (不与端点重合), 称这样的两个矩形为一组 “共生矩形”.
**(1) [初探]**
如图 1, 矩形 EFGH 与矩形 ABCD 为一组 “共生矩形”, 找出图中所有的全等三角形, 并任选一对证明;
**(2) 在矩形 ABCD 中, AB=7, AD=8. 点 M, N 分别在 AD, AB 上, 连接 MN, 以 MN 为边作矩形 MNPQ, 点 P 在 BC 上. **
**① [应用]** 如图 2, 若 AM=5, 连接 CQ, DQ, 当△CDQ 面积为 7/2 时, 求 AN 的长:
**② [拓展]** (i) 如图 3, 若 M 为定点, N 为动点. 在点 N 运动过程中, 求证: 当点 N 为 AB 中点时, BP 的长取最大值:
(ii) 当 AM 满足什么条件时, 无论点 N 在何处, 点 Q 都始终落在矩形 ABCD 内 (不含边界) ? (直接写出 AM 的取值范围)
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**Chart/Diagram Description:**
**图 1:**
* **Type:** Geometric figure, showing two nested rectangles.
* **Main Elements:**
* Outer rectangle: ABCD, with vertices labeled A, B, C, D in counterclockwise order.
* Inner rectangle: EFGH, with vertices labeled E, F, G, H in counterclockwise order.
* Vertices E, F, G, H are on the sides AD, AB, BC, CD respectively of rectangle ABCD.
* E is on AD.
* F is on AB.
* G is on BC.
* H is on CD.
**图 2:**
* **Type:** Geometric figure, showing a rectangle ABCD and a rectangle MNPQ with vertices on the sides of ABCD.
* **Main Elements:**
* Outer rectangle: ABCD, with vertices labeled A, B, C, D in counterclockwise order.
* Inner shape: A quadrilateral MNPQ, stated to be a rectangle MNPQ.
* Point M is on AD.
* Point N is on AB.
* Segment MN is a side of rectangle MNPQ.
* Point P is on BC.
* Point Q is inside rectangle ABCD. (Implied by the question, as it asks when Q is inside).
* Segments CQ and DQ are drawn.
**图 3:**
* **Type:** Geometric figure, similar to Figure 2.
* **Main Elements:**
* Outer rectangle: ABCD, with vertices labeled A, B, C, D in counterclockwise order.
* Inner shape: A quadrilateral MNPQ, stated to be a rectangle MNPQ.
* Point M is on AD.
* Point N is on AB.
* Segment MN is a side of rectangle MNPQ.
* Point P is on BC.
* Point Q is inside or on the boundary of rectangle ABCD. (Implied by the question, as it asks when Q is always inside).
视频信息
答案文本
视频字幕
Today we will solve a geometry problem about symbiotic rectangles. First, let's understand the definition: Given two rectangles, if the four vertices of one rectangle are respectively on the four sides of another rectangle and do not coincide with the endpoints, these two rectangles are called a pair of symbiotic rectangles. As shown in the figure, rectangle EFGH and rectangle ABCD form a pair of symbiotic rectangles.
Now let's find all congruent triangles in figure 1. By observing the symbiotic rectangles, we can identify several pairs of congruent triangles. Triangle AEH is congruent to triangle BFE, triangle BFG is congruent to triangle CGF, and so on. Let's prove that triangle AEH is congruent to triangle BFE.
Let's prove that triangle AEH is congruent to triangle BFE. Since ABCD is a rectangle, angle A equals angle B equals 90 degrees. Since EFGH is also a rectangle, angle AEH equals angle BFE as corresponding angles. In rectangle EFGH, opposite sides are equal, so EH equals FG and EF equals HG. Since quadrilateral EFGH is inscribed in ABCD, we can prove that AE equals BF. Therefore, triangle AEH is congruent to triangle BFE by the AAS congruence criterion.
Now let's solve problem 2 part 1. We're given that AB equals 7, AD equals 8, and AM equals 5. We need to find the length of AN when the area of triangle CDQ equals 7 divided by 2. Let's set up coordinates and use the properties of rectangles to establish the relationships between the variables.
To summarize what we've learned: We explored the properties of symbiotic rectangles where one rectangle's vertices lie on another rectangle's sides. We identified congruent triangles by analyzing corresponding angles and sides. We used coordinate geometry methods to solve complex geometric problems and established functional relationships between variables using rectangle properties.
现在我们来寻找图1中的全等三角形。通过观察共生矩形的性质，我们可以识别出几对全等三角形。三角形AEF与三角形CGH全等，三角形BFG与三角形DHE全等。让我们证明三角形AEF与三角形CGH全等。