The sides AB, BC and AC of a `DeltaABC` are 12 cm, 8 cm and 10 cm respectively. A circle is inscribed in the triangle touching AB, BC and AC at D, E and F, respectively. The difference between the lengths of AD and CE is:

Options

A. a)4 cm
B. b)5 cm
C. c)3 cm
D. d)2cm

Answer

To solve the problem, we need to find the difference between the lengths of segments AD and CE in triangle ABC, where a circle is inscribed. The sides of the triangle are given as follows:

- AB = 12 cm
- BC = 8 cm
- AC = 10 cm

### Step 1: Define the segments
Let:
- AD = x
- DB = y
- EC = z
- FC = w

From the properties of tangents drawn from an external point to a circle, we know:
- AD = AF = x
- DB = BE = y
- EC = EF = z
- FC = FD = w

### Step 2: Establish relationships
From the triangle sides, we can establish the following relationships:
- For side AB: AD + DB = AB ⇒ x + y = 12
- For side BC: BE + EC = BC ⇒ y + z = 8
- For side AC: AF + FC = AC ⇒ x + w = 10

### Step 3: Express the lengths in terms of x, y, and z
From the equations:
1. \( x + y = 12 \) (1)
2. \( y + z = 8 \) (2)
3. \( x + w = 10 \) (3)

### Step 4: Solve for y and z
From equation (1), we can express y in terms of x:
\[ y = 12 - x \]

Substituting y in equation (2):
\[ (12 - x) + z = 8 \]
\[ z = 8 - (12 - x) \]
\[ z = x - 4 \] (4)

### Step 5: Find the difference AD - CE
We need to find \( AD - CE \):
\[ AD - CE = x - z \]
Substituting z from equation (4):
\[ AD - CE = x - (x - 4) \]
\[ AD - CE = 4 \]

### Conclusion
The difference between the lengths of AD and CE is **4 cm**.