Okay, let's solve the last problem: Finding the value of AM without using vectors.
**Problem Statement:**
In ΔABC, ∠BAC = 60°, M is the midpoint of BC. Construct equilateral triangle ΔBCD downwards on side BC. Connect AM, AD, MD. The distance from M to AD is 1, and the area of ΔAMD is 3.6. Find the value of AM.
**Solution:**
1. **Use the Area Information to find AD:**
* The area of a triangle is given by (1/2) * base * height.
* For ΔAMD, we are given the area is 3.6 and the height (distance) from M to the base AD is 1.
* Area(ΔAMD) = (1/2) * AD * (distance from M to AD)
* 3.6 = (1/2) * AD * 1
* 3.6 = AD / 2
* AD = 3.6 * 2 = 7.2
2. **Relate AD and AM using Geometric Properties:**
* This configuration (ΔABC with ∠BAC=60°, M midpoint of BC, equilateral ΔBCD constructed on BC) has a known property that relates AD and AM. Let's derive it.
* Let M be the origin (0,0). Let BC lie on the x-axis. Let C = (c, 0) and B = (-c, 0). So BC = 2c and M is the midpoint.
* Since ΔBCD is equilateral and constructed downwards, the coordinates of D are (0, -c√3).
* The length MD = √[(0-0)² + (-c√3 - 0)²] = √(3c²) = c√3. Note that $MD^2 = 3c^2$. Also, $MD = \frac{\sqrt{3}}{2} (2c) = \frac{\sqrt{3}}{2} BC$.
* Let the coordinates of A be (x, y). Then $AM^2 = x^2 + y^2$.
* The distance AD is given by $AD^2 = (x-0)^2 + (y - (-c\sqrt{3}))^2 = x^2 + (y+c\sqrt{3})^2 = x^2 + y^2 + 2yc\sqrt{3} + 3c^2$.
* Substitute $AM^2 = x^2+y^2$ and $MD^2 = 3c^2$:
$AD^2 = AM^2 + MD^2 + 2(c\sqrt{3})y = AM^2 + MD^2 + 2 \cdot MD \cdot y$
* Now we need to use the condition ∠BAC = 60°. Let h be the altitude from A to BC. In our coordinate system, h = |y|. The area of ΔABC, denoted S, is S = (1/2) * BC * h = (1/2) * (2c) * |y| = c|y|.
* Also, by the Law of Cosines in ΔABC:
$BC^2 = AB^2 + AC^2 - 2 AB \cdot AC \cos(60°)$
$(2c)^2 = AB^2 + AC^2 - AB \cdot AC$
* By Apollonius' Theorem on median AM in ΔABC:
$AB^2 + AC^2 = 2(AM^2 + BM^2) = 2(AM^2 + c^2)$
* Substituting this into the Law of Cosines equation:
$4c^2 = 2(AM^2 + c^2) - AB \cdot AC$
$AB \cdot AC = 2(AM^2 + c^2) - 4c^2 = 2AM^2 - 2c^2 = 2(AM^2 - c^2)$
* The area of ΔABC is also given by $S = (1/2) AB \cdot AC \sin(60°) = (1/2) [2(AM^2 - c^2)] (\sqrt{3}/2) = \frac{\sqrt{3}}{2}(AM^2 - c^2)$.
* Comparing the two expressions for the area S:
$c|y| = \frac{\sqrt{3}}{2}(AM^2 - c^2)$
$|y| = \frac{\sqrt{3}}{2c}(AM^2 - c^2)$
* From the diagram, A is generally "above" BC, and D is constructed "below", meaning y (the y-coordinate of A relative to M) is likely positive. Let's assume y > 0. Then $y = \frac{\sqrt{3}}{2c}(AM^2 - c^2)$.
* Substitute this value of y into the expression for AD²:
$AD^2 = AM^2 + MD^2 + 2 \cdot MD \cdot y$
$AD^2 = AM^2 + (c\sqrt{3})^2 + 2 (c\sqrt{3}) \left[ \frac{\sqrt{3}}{2c}(AM^2 - c^2) \right]$
$AD^2 = AM^2 + 3c^2 + \frac{2 c \sqrt{3} \cdot \sqrt{3}}{2c}(AM^2 - c^2)$
$AD^2 = AM^2 + 3c^2 + 3(AM^2 - c^2)$
$AD^2 = AM^2 + 3c^2 + 3AM^2 - 3c^2$
$AD^2 = 4AM^2$
* Taking the square root (lengths are positive):
$AD = 2AM$
3. **Calculate AM:**
* We found AD = 7.2 and derived the relationship AD = 2AM for this specific geometric configuration (∠BAC=60°, M midpoint BC, equilateral ΔBCD outwards).
* 7.2 = 2 * AM
* AM = 7.2 / 2
* AM = 3.6
**Answer:**
The value of AM is 3.6.作者: wolfydw 时间: 2025-4-24 20:33
首先，这是初二全等三角形的题目，属于全等三角形的手拉手模型。其次，你得给我发红包。作者: reg01 时间: 2025-4-24 20:50
AEC = ADB 因为 角CAE=角BAD. 两个角各自的两条边又等长，作者: Roman 时间: 2025-4-24 20:51
这是八上的题，我家孩子虽然是七年级，不过应该没什么问题。作者: 方块李 时间: 2025-4-25 08:50
**20．（11 分）答案与推导**
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### 一、【模型构建】（图①）
1. **证明全等**
\[
\triangle ABD \cong \triangle ACE
\]
