Given, the angles A, B and C of `triangleABC`. Let M be the mid-point of segment AB and let D be the foot of the bisector of `angleC`. Find the ratio of `(Area Of triangleCDM)/(Area of triangleABC)`

Answer

To find the ratio of the area of triangle \(CDM\) to the area of triangle \(ABC\), we can follow these steps:

### Step 1: Understand the Configuration
Let \(A\), \(B\), and \(C\) be the vertices of triangle \(ABC\). Let \(M\) be the midpoint of segment \(AB\) and \(D\) be the foot of the bisector of angle \(C\).

### Step 2: Area of Triangle \(CDM\)
The area of triangle \(CDM\) can be expressed as:
\[ \text{Area of } \triangle CDM = \frac{1}{2} \times DM \times CT \]
where \(DM\) is the base and \(CT\) is the height from point \(C\) to line \(DM\).

### Step 3: Area of Triangle \(ABC\)
The area of triangle \(ABC\) can be expressed as:
\[ \text{Area of } \triangle ABC = \frac{1}{2} \times AB \times CT \]
where \(AB\) is the base and \(CT\) is the height from point \(C\) to line \(AB\).

### Step 4: Set Up the Ratio
Now, we can set up the ratio of the areas:
\[ \frac{\text{Area of } \triangle CDM}{\text{Area of } \triangle ABC} = \frac{\frac{1}{2} \times DM \times CT}{\frac{1}{2} \times AB \times CT} \]
The \(CT\) cancels out:
\[ = \frac{DM}{AB} \]

### Step 5: Find \(DM\)
Using the angle bisector theorem, we know that:
\[ \frac{AC}{BC} = \frac{AD}{DB} \]
Let \(AC = b\), \(BC = a\), and \(AB = c\). Then:
\[ \frac{b}{a} = \frac{AD}{DB} \]
Let \(AD = x\) and \(DB = y\). Thus:
\[ \frac{b}{a} = \frac{x}{y} \implies b \cdot y = a \cdot x \]
From this, we can express \(AD\) and \(DB\) in terms of \(c\):
\[ AD + DB = AB \implies x + y = c \]
Substituting \(y = \frac{b}{a}x\) into \(x + y = c\):
\[ x + \frac{b}{a}x = c \implies x\left(1 + \frac{b}{a}\right) = c \implies x = \frac{c}{1 + \frac{b}{a}} = \frac{ac}{a + b} \]
Thus, \(AD = \frac{ac}{a + b}\) and \(DB = \frac{bc}{a + b}\).

### Step 6: Calculate \(DM\)
Since \(M\) is the midpoint of \(AB\):
\[ DM = DB - MB \]
where \(MB = \frac{c}{2}\):
\[ DM = \frac{bc}{a + b} - \frac{c}{2} \]
This can be simplified further, but we can also express it in terms of \(c\):
\[ DM = \frac{bc - \frac{c(a + b)}{2}}{a + b} = \frac{2bc - c(a + b)}{2(a + b)} = \frac{c(b - \frac{a + b}{2})}{a + b} \]

### Step 7: Final Ratio
Now substituting back into our ratio:
\[ \frac{DM}{AB} = \frac{\frac{c(b - \frac{a + b}{2})}{a + b}}{c} = \frac{b - \frac{a + b}{2}}{a + b} \]
This simplifies to:
\[ \frac{2b - (a + b)}{2(a + b)} = \frac{b - a}{2(a + b)} \]

### Conclusion
Thus, the final ratio of the areas is:
\[ \frac{\text{Area of } \triangle CDM}{\text{Area of } \triangle ABC} = \frac{b - a}{2(a + b)} \]