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: 
\[ 
\boxed{5.625} 
\] 
  | 
 
  
 
 |   
 
 |