根据附件图片内容，讲解 等分面积模型---Here is the extraction of the content from the image:
**Title:** 模型 52 “等分面积”模型 (Model 52 "Area Division" Model)
**Basic Model:**
* **Title:** 基础模型 (Basic Model)
* **Diagram:**
* Type: Geometric figure (Triangle)
* Main Elements: Triangle ABC with vertices A, B, C. Point D is on side BC. Line segment AD is drawn.
* **Conditions:** 在△ABC中,过边BC上一点D作一条直线将三角形面积平分 (In △ABC, draw a line through point D on side BC to divide the triangle's area in half)
* **Construction Method:**
* Method 1: 作法一: 取BC的中点E, 连接AE,AD,过点E作EF//AD交AC于点F,过D,F两点作直线,交AE于点O (Method 1: Take the midpoint E of BC, connect AE, AD, draw EF//AD through point E intersecting AC at point F, draw a line through points D, F intersecting AE at point O)
* Diagram: Triangle ABC, point D on BC, E is the midpoint of BC, AD, AE, EF are drawn. Line DF is drawn intersecting AE at O. EF is parallel to AD. Shaded regions are △AOF and △DOE.
* Method 2: 作法二: 连接AD,过点B作BE//AD,交CA延长线于点E,连接AE,取CD的中点F,过D,F两点作直线 (Method 2: Connect AD, draw BE//AD through point B intersecting the extension of CA at point E, connect AE, take the midpoint F of CD, draw a line through points D, F)
* Diagram: Triangle ABC, point D on BC, AD, BE, AE are drawn. E is on the extension of CA. BE is parallel to AD. F is the midpoint of CD. Line DF is drawn intersecting AB at O. Shaded regions are △ADF and △BDF.
* **Conclusion:**
* Method 1:
1. S△AEF=S△EFD; (Area(triangle AEF) = Area(triangle EFD))
2. S△AOF=S△DOE; (Area(triangle AOF) = Area(triangle DOE))
3. S△DFC=S四边形ABDF (Area(triangle DFC) = Area(quadrilateral ABDF))
* Method 2:
1. S△ABD=S△EAD; (Area(triangle ABD) = Area(triangle EAD))
2. S△AOD=S△BOD; (Area(triangle AOD) = Area(triangle BOD))
3. S△ADF=S△CDF (Area(triangle ADF) = Area(triangle CDF))
**Model Tips:**
* **Title:** 模型巧记 (Model Tips)
* 1. 解决三角形和不规则四边形的面积平分问题常会用到下面两个重要结论: (Solving area division problems for triangles and irregular quadrilaterals often uses the following two important conclusions:)
* (1) 同底等高的两个三角形面积相等,常作平行构造面积相等的两个三角形转化面积; ((1) Two triangles with the same base and equal height have equal areas, often use parallel lines to construct two triangles with equal areas for area transformation;)
* (2) 中线等分面积 ((2) A median divides the area in half)
* 2. 解决规则图形的面积平分问题,可借助特殊图形的对称性,经过其中心的直线可将图形面积等分. (Solving area division problems for regular figures can utilize the symmetry of special figures; a line passing through the center can divide the figure's area in half.)
**Model Extension (Advanced Study):**
* **Title:** 悬考延伸 (Model Extension)
* **Description:** 同学们还可尝试利用作法二的方式进行结论证明哦! (Students can also try to prove the conclusions using Method 2!)
**Conclusion Analysis:**
* **Title:** 【结论分析】 (Conclusion Analysis)
* **Content:** 作法一结论:1. S△AEF=S△EFD;2. S△AOF=S△DOE;3. S△DFC=S四边形ABDF
证明:: EF//AD, ∴ S△AEF=S△EFD (结论1), ∴ S△AEF=S△AOF+S△EOF, S△EFD=S△DOE+S△EOF, ∴ S△AOF=S△DOE (结论2). ∴ E是BC的中点, ∴ S△ABE=S△AEC, ∴ S△ABE=S四边形ABDO+S△DOE, S△AEC=S四边形CEOF+S△AOF, ∴ S四边形ABDO+S△DOE=S四边形CEOF+S△AOF, 将结论2 S△AOF=S△DOE代入, 得S四边形ABDO+S△AOF=S四边形CEOF+S△AOF, S四边形ABDO=S四边形CEOF, S△ABC=S△ABE+S△AEC=2S△ABE=2(S四边形ABDO+S△DOE). S△DFC=S△ABC - S四边形ABDF, S四边形ABDF = S四边形ABDO + S△DOF? (text unclear) ... 证明了结论3. (证明:: EF//AD, ∴ Area(triangle AEF)=Area(triangle EFD) (Conclusion 1), ∴ Area(triangle AEF)=Area(triangle AOF)+Area(triangle EOF), Area(triangle EFD)=Area(triangle DOE)+Area(triangle EOF), ∴ Area(triangle AOF)=Area(triangle DOE) (Conclusion 2). ∴ E is the midpoint of BC, ∴ Area(triangle ABE)=Area(triangle AEC), ∴ Area(triangle ABE)=Area(quadrilateral ABDO)+Area(triangle DOE), Area(triangle AEC)=Area(quadrilateral CEOF)+Area(triangle AOF), ∴ Area(quadrilateral ABDO)+Area(triangle DOE)=Area(quadrilateral CEOF)+Area(triangle AOF), Substitute conclusion 2 Area(triangle AOF)=Area(triangle DOE), we get Area(quadrilateral ABDO)+Area(triangle AOF)=Area(quadrilateral CEOF)+Area(triangle AOF), Area(quadrilateral ABDO)=Area(quadrilateral CEOF), Area(triangle ABC)=Area(triangle ABE)+Area(triangle AEC)=2*Area(triangle ABE)=2*(Area(quadrilateral ABDO)+Area(triangle DOE)). Area(triangle DFC)=Area(triangle ABC) - Area(quadrilateral ABDF)... Proved conclusion 3.)
**Extended Model:**
* **Title:** 拓展模型 (Extended Model)
* **Conditions:** 条件:在四边形ABCD中,过顶点A作一条直线将四边形ABCD的面积平分 (Conditions: In quadrilateral ABCD, draw a line through vertex A to divide the area of quadrilateral ABCD in half)
* **Diagram (Initial):**
* Type: Geometric figure (Quadrilateral)
* Main Elements: Quadrilateral ABCD.
* **Construction Method:** 作法:如图,连接AC,过点D作DE//AC,交BC的延长线于点E,连接AE,取BE的中点F,过A,F两点作直线 (Construction Method: As shown, connect AC, draw DE//AC through point D intersecting the extension of BC at point E, connect AE, take the midpoint F of BE, draw a line through points A, F)
* **Diagram (Construction):**
* Type: Geometric figure (Quadrilateral with construction)
* Main Elements: Quadrilateral ABCD. Diagonal AC. Line DE drawn through D parallel to AC, intersecting extended BC at E. Line AE. F is the midpoint of BE. Line AF is drawn.
* **Conclusion:** 结论:S△ABF=S四边形AFCD (Conclusion: Area(triangle ABF) = Area(quadrilateral AFCD))
**Example 1:**
* **Title:** 例1 一题多解 (Example 1 - Multiple Solutions)
* **Question Stem:** 如图,在△ABC中,AB=AC=5,BC=6,D为BC边上一点,且BD=1,在AC边上找一点E,使得DE平分△ABC的面积,则CE的长为___. (As shown in the figure, in △ABC, AB=AC=5, BC=6, D is a point on side BC, and BD=1. Find a point E on side AC such that DE divides the area of △ABC in half, then the length of CE is ___.)
* **Diagram:**
* Type: Geometric figure (Triangle with line segment)
* Main Elements: Triangle ABC. Point D is on BC, with BD=1. Point E is on AC. Line segment DE is drawn. AB=5, AC=5, BC=6.
* Label: 例1题图 (Example 1 Problem Diagram)
* **Model Idea Bubble:** 过边上一点作一条直线将三角形的面积平分,故为“等分面积”模型 (Draw a line through a point on a side to divide the triangle's area in half, hence it is the "Area Division" model)
* **Step-by-step thinking process:**
* **Step 1:** 第一步: 依据特征找模型 (Step 1: Identify the model based on characteristics)
* Content: 特征1: 是否存在三角形 △ABC (Characteristic 1: Is there a triangle △ABC); 特征2: 在三角形的边上是否存在一点且过该点所在的直线平分三角形的面积 BC边上的点D (Characteristic 2: Is there a point on the side of the triangle and does the line passing through this point divide the area of the triangle in half? Point D on side BC)
* **Step 2:** 第二步: 抽象模型 (Step 2: Abstract the model)
* Diagram: Triangle ABC with point D on BC.
* **Step 3:** 第三步: 用模型 (Step 3: Use the model)
* Description: 取BC的中点F,连接AF,AD,过点F作AD的平行线,交CA于点E,连接DE. (Take the midpoint F of BC, connect AF, AD, draw a line parallel to AD through point F, intersecting CA at point E, connect DE.)
* Diagram: Triangle ABC. D on BC, F is the midpoint of BC. AD, AF, DE, EF are drawn. EF is parallel to AD. E is on AC.
* Formula/Equality: S四边形ABDE = S△CDE (Area(quadrilateral ABDE) = Area(triangle CDE))