SOLUTION: Point D is the midpoint of median AM of triangle ABC. Point E is the midpoint of AB, and point T is the intersection of BD and ME. Find the area of triangle DMT if [ABC] = 180.

Question 1192682: Point D is the midpoint of median AM of triangle ABC. Point E is the midpoint of AB, and point T is the intersection of BD and ME. Find the area of triangle DMT if [ABC] = 180.
Answer by textot(100) (Show Source): You can put this solution on YOUR website!
We aim to solve this problem step-by-step by analyzing the geometry of the triangle and using properties of medians and midpoints.
---
### Step 1: Notation and Breakdown
- Triangle \( \triangle ABC \) has area \([ABC] = 180\).
- \( M \) is the midpoint of \( BC \), and \( AM \) is a median.
- \( D \) is the midpoint of \( AM \).
- \( E \) is the midpoint of \( AB \).
- \( T \) is the intersection of \( BD \) and \( ME \).
- We are tasked to find \([DMT]\), the area of triangle \( \triangle DMT \).
---
### Step 2: Key Geometry Properties
1. **Area Split by Median**:
Median \( AM \) divides \( \triangle ABC \) into two triangles (\( \triangle ABM \) and \( \triangle ACM \)) of equal area. Thus:
\[
[ABM] = [ACM] = \frac{1}{2} \cdot [ABC] = \frac{1}{2} \cdot 180 = 90.
\]
2. **Midpoint Property**:
\( D \) is the midpoint of \( AM \), and \( E \) is the midpoint of \( AB \). Both \( D \) and \( E \) further divide areas into smaller equal parts.
---
### Step 3: Triangle Area Division via Proportionality
Using the midpoint theorem:
1. \( D \) divides \( AM \) into two equal segments, so any triangle formed with \( D \) has an area \( \frac{1}{2} \) of the corresponding triangle with \( A \).
2. \( ME \) and \( BD \) are constructed from midpoints. The intersection \( T \) divides \( \triangle DME \) into smaller proportional areas.
---
### Step 4: Area of \( \triangle DMT \)
To find the area of \( \triangle DMT \), we rely on symmetry and proportionality:
1. The area of \( \triangle DME \) is proportional to \( \frac{1}{8} \) of \( [ABC] \), because both \( D \) and \( E \) divide their respective triangles into halves, and intersecting midlines divide the triangle into smaller, equally proportional regions.
2. The triangle \( \triangle DMT \) is one-fourth of \( \triangle DME \) because \( T \) divides \( \triangle DME \) into four equal parts.
Thus:
\[
[DME] = \frac{1}{8} \cdot [ABC] = \frac{1}{8} \cdot 180 = 22.5,
\]
and:
\[
[DMT] = \frac{1}{4} \cdot [DME] = \frac{1}{4} \cdot 22.5 = 5.625.
\]
---
### Final Answer:
The area of \( \triangle DMT \) is:
\[
\boxed{5.625}
\]
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