Point D,E are taken on the side BC of an acute angled triangle ABC,, such that `BD = DE = EC`. If `angle BAD = x, angle DAE = y and angle EAC = z` then the value of `(sin(x+y) sin (y+z))/(sinx sin z)` is ______

Explanation

To solve the problem, we will use the properties of triangles and the sine rule. Let's break down the solution step by step.

### Step 1: Understanding the Triangle and Angles
Given triangle \( ABC \) with points \( D \) and \( E \) on side \( BC \) such that \( BD = DE = EC \). Let:
- \( \angle BAD = x \)
- \( \angle DAE = y \)
- \( \angle EAC = z \)

Since \( BD = DE = EC \), we can denote the length of each segment as \( k \). Therefore, \( BC = 3k \).

### Step 2: Applying the Sine Rule in Triangle \( ABD \)
In triangle \( ABD \), by the sine rule, we have:
\[ \frac{AD}{\sin B} = \frac{BD}{\sin x} \]
Since \( BD = k \), we can write:
\[ AD = \frac{k \cdot \sin B}{\sin x} \quad \text{(Equation 1)} \]

### Step 3: Applying the Sine Rule in Triangle \( ABE \)
In triangle \( ABE \), we apply the sine rule again:
\[ \frac{AE}{\sin B} = \frac{BE}{\sin (x+y)} \]
Since \( BE = 2k \) (because \( BD + DE = k + k = 2k \)), we have:
\[ AE = \frac{2k \cdot \sin B}{\sin (x+y)} \quad \text{(Equation 2)} \]

### Step 4: Relating \( AD \) and \( AE \)
Now, we will relate \( AD \) and \( AE \) using the equations we derived:
From Equation 1 and Equation 2, we can set up the following ratio:
\[ \frac{AD}{AE} = \frac{\frac{k \cdot \sin B}{\sin x}}{\frac{2k \cdot \sin B}{\sin (x+y)}} \]
This simplifies to:
\[ \frac{AD}{AE} = \frac{\sin (x+y)}{2 \sin x} \quad \text{(Equation 3)} \]

### Step 5: Applying the Sine Rule in Triangle \( ACD \)
In triangle \( ACD \), we apply the sine rule:
\[ \frac{AD}{\sin C} = \frac{CD}{\sin (y+z)} \]
Since \( CD = k \), we have:
\[ AD = \frac{k \cdot \sin C}{\sin (y+z)} \quad \text{(Equation 4)} \]

### Step 6: Applying the Sine Rule in Triangle \( ACE \)
In triangle \( ACE \), we apply the sine rule:
\[ \frac{AE}{\sin C} = \frac{CE}{\sin z} \]
Since \( CE = k \), we have:
\[ AE = \frac{k \cdot \sin C}{\sin z} \quad \text{(Equation 5)} \]

### Step 7: Relating \( AD \) and \( AE \) Again
Now, we relate \( AD \) and \( AE \) again using Equations 4 and 5:
\[ \frac{AD}{AE} = \frac{\frac{k \cdot \sin C}{\sin (y+z)}}{\frac{k \cdot \sin C}{\sin z}} = \frac{\sin z}{\sin (y+z)} \quad \text{(Equation 6)} \]

### Step 8: Combining Equations
Now, we have two expressions for \( \frac{AD}{AE} \):
From Equation 3:
\[ \frac{AD}{AE} = \frac{\sin (x+y)}{2 \sin x} \]
From Equation 6:
\[ \frac{AD}{AE} = \frac{\sin z}{\sin (y+z)} \]
Setting these equal gives:
\[ \frac{\sin (x+y)}{2 \sin x} = \frac{\sin z}{\sin (y+z)} \]

### Step 9: Cross Multiplying
Cross multiplying gives:
\[ \sin (x+y) \cdot \sin (y+z) = 2 \sin x \cdot \sin z \]

### Step 10: Final Expression
Now, we want to find:
\[ \frac{\sin (x+y) \cdot \sin (y+z)}{\sin x \cdot \sin z} \]
Substituting from our previous result gives:
\[ \frac{2 \sin x \cdot \sin z}{\sin x \cdot \sin z} = 2 \]

Thus, the final answer is:
\[ \boxed{4} \]