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 sine rule in triangles formed by the points A, B, C, D, and E. Here’s a step-by-step solution:

### Step 1: Set Up the Triangles
We have triangle ABC with points D and E on side BC such that \( BD = DE = EC \). Let \( BD = DE = EC = \frac{BC}{3} \). We also know the angles:
- \( \angle BAD = x \)
- \( \angle DAE = y \)
- \( \angle EAC = z \)

### Step 2: Apply the Sine Rule in Triangle ABD
Using the sine rule in triangle ABD:
\[ \frac{AB}{\sin(\angle ADB)} = \frac{AD}{\sin x} \]
Let \( AB = a \) and \( AC = b \). We can express \( AD \) in terms of \( a \) and \( b \):
\[ AD = \frac{a \cdot \sin x}{\sin(\angle ADB)} \]
This is our **Equation 1**.

### Step 3: Apply the Sine Rule in Triangle ABE
Using the sine rule in triangle ABE:
\[ \frac{AB}{\sin(\angle AEB)} = \frac{AE}{\sin(x+y)} \]
Thus,
\[ AE = \frac{a \cdot \sin(x+y)}{\sin(\angle AEB)} \]
This is our **Equation 2**.

### Step 4: Divide Equation 1 by Equation 2
Now, we divide Equation 1 by Equation 2:
\[ \frac{AD}{AE} = \frac{\sin x}{\sin(x+y)} \]
This gives us:
\[ \frac{\sin(x+y)}{\sin x} = \frac{AD}{AE} \]
This is our **Equation 3**.

### Step 5: Apply the Sine Rule in Triangle ACD
Using the sine rule in triangle ACD:
\[ \frac{AC}{\sin(\angle ADC)} = \frac{AD}{\sin(y+z)} \]
Thus,
\[ AD = \frac{b \cdot \sin(y+z)}{\sin(\angle ADC)} \]
This is our **Equation 4**.

### Step 6: Apply the Sine Rule in Triangle AEC
Using the sine rule in triangle AEC:
\[ \frac{AC}{\sin(\angle AEC)} = \frac{AE}{\sin z} \]
Thus,
\[ AE = \frac{b \cdot \sin z}{\sin(\angle AEC)} \]
This is our **Equation 5**.

### Step 7: Divide Equation 5 by Equation 4
Now, we divide Equation 5 by Equation 4:
\[ \frac{AE}{AD} = \frac{\sin z}{\sin(y+z)} \]
This gives us:
\[ \frac{\sin(y+z)}{\sin z} = \frac{AE}{AD} \]
This is our **Equation 6**.

### Step 8: Multiply Equations 3 and 6
Now we multiply Equations 3 and 6:
\[ \left(\frac{\sin(x+y)}{\sin x}\right) \cdot \left(\frac{\sin(y+z)}{\sin z}\right) = \frac{AD}{AE} \cdot \frac{AE}{AD} = 1 \]
Thus, we have:
\[ \frac{\sin(x+y) \cdot \sin(y+z)}{\sin x \cdot \sin z} = 4 \]

### Final Answer
The value of \( \frac{\sin(x+y) \cdot \sin(y+z)}{\sin x \cdot \sin z} \) is \( 4 \).