根据附件图片内容，讲解“等分面积模型”---Here is the extracted content from the image:
**模型 52 “等分面积”模型 (Model 52 "Area Bisection" Model)**
**模型展现 (Model Presentation)**
**基础模型 (Basic Model)**
| 图示 (Diagram) | 条件 (Conditions) | 作法一 (Construction Method 1) | 作法二 (Construction Method 2) |
| :------------- | :-------------------------------------------------------------------------------- | :------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | :-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| | 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 which bisects the area of the triangle) | 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 (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 and F, intersecting AE at point O) | 连接AD, 过点B作BE//AD,交CA延长线于点E, 连接DE,交AB于点O,取CE的中点F,过D,F两点作直线 (Connect AD, draw BE // AD through point B intersecting CA extended at point E, connect DE, intersecting AB at point O, take the midpoint F of CE, draw a line through points D and F) |
| | | **Conclusions:** 1. $S_{\triangle AEF} = S_{\triangle EFD}$ 2. $S_{\triangle AOF} = S_{\triangle DOE}$ 3. $S_{\triangle DFC} = S_{\text{quadrilateral } ABDF}$ | **Conclusions:** 1. $S_{\triangle ABD} = S_{\triangle ADE}$ 2. $S_{\triangle AOF} = S_{\triangle DOB}$ 3. $S_{\triangle DFC} = S_{\text{quadrilateral } ABDF}$ |
**Diagram for Method 1:**
A triangle ABC with point D on BC. E is the midpoint of BC. Lines AE and AD are drawn. A line EF is drawn through E parallel to AD, with F on AC. A line DF is drawn, intersecting AE at O. Areas $\triangle ADF$ and $\triangle ADE$ are shaded. Areas $\triangle AOF$ and $\triangle DOE$ are shaded with a darker pattern. Dashed lines are AE and DF.
**Diagram for Method 2:**
A triangle ABC with point D on BC. Line AD. Line BE is drawn parallel to AD, with E on the extension of CA. Line DE is drawn. F is the midpoint of CE. Line DF is drawn. Areas $\triangle ABD$ and $\triangle ADE$ are shaded. Dashed lines are AD and BE.
**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: (To solve problems involving area bisection of triangles and irregular quadrilaterals, the following two important conclusions are often used:)
(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) Triangles with the same base and equal height have equal area, often constructing parallel lines to transform areas of two triangles with equal area;)
(2) 中线等分面积. ((2) The median bisects the area.)
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. (To solve area bisection problems for regular figures, one can utilize the symmetry of the special figure, and a line passing through its center of symmetry can bisect the area of the figure.)
**【结论分析】 (Conclusion Analysis)**
作法一结论: 1. $S_{\triangle AEF} = S_{\triangle EFD}$; 2. $S_{\triangle AOF} = S_{\triangle DOE}$; 3. $S_{\triangle DFC} = S_{\text{四边形} ABDF}$ (Method 1 Conclusions: ...)
证明: ∵EF//AD, ∴$S_{\triangle AEF} = S_{\triangle EFD}$ (结论1). ∵同底等高的两个三角形面积相等. (Proof: Because EF//AD, therefore $S_{\triangle AEF} = S_{\triangle EFD}$ (Conclusion 1). Because triangles with the same base and equal height have equal area.)
$S_{\triangle AEF} = S_{\triangle AOF} + S_{\triangle EOF}$, $S_{\triangle EFD} = S_{\triangle EOF} + S_{\triangle DOE}$, ∴$S_{\triangle AOF} = S_{\triangle DOE}$ (结论2). ( $S_{\triangle AEF} = S_{\triangle AOF} + S_{\triangle EOF}$, $S_{\triangle EFD} = S_{\triangle EOF} + S_{\triangle DOE}$, therefore $S_{\triangle AOF} = S_{\triangle DOE}$ (Conclusion 2).)
∵E是BC的中点, ∴$S_{\triangle ABE} = S_{\triangle AEC}$, $S_{\triangle ABC} = S_{\triangle ABE} + S_{\triangle AEC}$. (Because E is the midpoint of BC, therefore $S_{\triangle ABE} = S_{\triangle AEC}$, $S_{\triangle ABC} = S_{\triangle ABE} + S_{\triangle AEC}$.)
$S_{\triangle ABE} = S_{\text{四边形} ABDO} + S_{\triangle DOE}$, $S_{\triangle AEC} = S_{\text{四边形} CEOF} + S_{\triangle AOF}$. ( $S_{\triangle ABE} = S_{\text{quadrilateral } ABDO} + S_{\triangle DOE}$, $S_{\triangle AEC} = S_{\text{quadrilateral } CEOF} + S_{\triangle AOF}$.)
∵$S_{\triangle ABE} = S_{\triangle AEC}$, ∴$S_{\text{四边形} ABDO} + S_{\triangle DOE} = S_{\text{四边形} CEOF} + S_{\triangle AOF}$. (Because $S_{\triangle ABE} = S_{\triangle AEC}$, therefore $S_{\text{quadrilateral } ABDO} + S_{\triangle DOE} = S_{\text{quadrilateral } CEOF} + S_{\triangle AOF}$.)
∵$S_{\triangle AOF} = S_{\triangle DOE}$, ∴$S_{\text{四边形} ABDO} = S_{\text{四边形} CEOF}$. (Because $S_{\triangle AOF} = S_{\triangle DOE}$, therefore $S_{\text{quadrilateral } ABDO} = S_{\text{quadrilateral } CEOF}$.)
$S_{\text{四边形} ABDO} + S_{\triangle DOE} = S_{\text{四边形} CEOF} + S_{\triangle AOF}$, 即$S_{\triangle DFC} = S_{\text{四边形} ABDF}$. ( $S_{\text{quadrilateral } ABDO} + S_{\triangle DOE} = S_{\text{quadrilateral } CEOF} + S_{\triangle AOF}$, which means $S_{\triangle DFC} = S_{\text{quadrilateral } ABDF}$.)
(结论3). ((Conclusion 3).)
**思孝延伸 (Ponder and Extend)**
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!)
**拓展模型 (Extended Model)**
| 条件 (Conditions) | 作法 (Construction Method) | 结论 (Conclusion) |
| :------------------------------------------------------------------- | :------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | :--------------------------------------------------- |
| 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 (In quadrilateral ABCD, draw a line through vertex A which bisects the area of quadrilateral ABCD) | 如图,连接AC,过点D作DE//AC,交BC的延长线于点E,连接AE, 交CD于点O,取BE的中点F,过A,F两点作直线 (As shown in the diagram, connect AC, draw DE // AC through point D intersecting the extension of BC at point E, connect AE, intersecting CD at point O, take the midpoint F of BE, draw a line through points A and F) | $S_{\triangle ABF} = S_{\text{quadrilateral } AFCD}$ |
**Diagram for Extended Model:**
A quadrilateral ABCD. Diagonal AC. A line DE is drawn through D parallel to AC, with E on the extension of BC. Line AE is drawn. F is the midpoint of BE. A line AF is drawn.
**例 1 一题多解 (Example 1 - Multiple Solutions)**
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 $\triangle ABC$, AB=AC=5, BC=6, D is a point on BC, and BD=1. Find a point E on side AC such that line DE bisects the area of $\triangle ABC$. What is the length of CE?)
**Diagram for Example 1:**
Triangle ABC, AB=AC=5, BC=6. Point D is on BC, closer to B, BD=1.
**模型猜想 (Model Guess)**
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 which bisects the area of the triangle, hence it is an "Area Bisection" model)
**第一步: 依据特征找模型 (Step 1: Identify Features and Find the Model)**
特征1: 是否存在三角形 (Feature 1: Does a triangle exist?)
△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 (Feature 2: Is there a point on a side and does the line passing through this point bisect the area of the triangle?)
BC边上的点D (Point D on side BC)
**第二步: 抽象模型 (Step 2: Abstract the Model)**
Diagram: Triangle ABC, point D on BC, point E on AC. Line segment DE.
**第三步: 用模型 (Step 3: Apply the Model)**
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 through point F parallel to AD, intersecting CA at point E, connect DE.)
**Diagram for Step 3:**
Triangle ABC. D is on BC. F is the midpoint of BC. Lines AF and AD are drawn. A line FE is drawn through F parallel to AD, with E on AC. Line DE is drawn.
Equation shown: $S_{\text{quadrilateral } ABDE} = S_{\triangle CDE}$