In `DeltaABC,angleA=(angleB+angleC)/(3)` and `angleB:angleC=2:1`. Find the measure of each angle of `DeltaABC` and state the type of `DeltaABC`.

Answer

To solve the problem, we will follow these steps systematically:

### Step 1: Set up the equations based on the given information.
We know that:
1. \( \angle A = \frac{\angle B + \angle C}{3} \) (Equation 1)
2. The ratio \( \angle B : \angle C = 2 : 1 \) (Equation 2)

Let \( \angle C = x \). Then, from the ratio, we have:
- \( \angle B = 2x \)

### Step 2: Use the angle sum property of a triangle.
The sum of the angles in a triangle is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \quad (Equation 3) \]

Substituting \( \angle B \) and \( \angle C \) from Equation 2 into Equation 3:
\[ \angle A + 2x + x = 180^\circ \]
This simplifies to:
\[ \angle A + 3x = 180^\circ \quad (Equation 4) \]

### Step 3: Substitute \( \angle A \) from Equation 1 into Equation 4.
From Equation 1, we have:
\[ \angle A = \frac{(2x + x)}{3} = \frac{3x}{3} = x \]

Now substitute \( \angle A \) in Equation 4:
\[ x + 3x = 180^\circ \]
This simplifies to:
\[ 4x = 180^\circ \]

### Step 4: Solve for \( x \).
Dividing both sides by 4:
\[ x = \frac{180^\circ}{4} = 45^\circ \]

### Step 5: Find \( \angle B \) and \( \angle C \).
Now that we have \( \angle C = x = 45^\circ \):
\[ \angle B = 2x = 2 \times 45^\circ = 90^\circ \]

### Step 6: Find \( \angle A \).
Using \( \angle A = x \):
\[ \angle A = 45^\circ \]

### Summary of the angles:
- \( \angle A = 45^\circ \)
- \( \angle B = 90^\circ \)
- \( \angle C = 45^\circ \)

### Step 7: Determine the type of triangle.
Since \( \angle B = 90^\circ \) and \( \angle A = \angle C = 45^\circ \), triangle ABC is a **right isosceles triangle**.

### Final Answer:
- \( \angle A = 45^\circ \)
- \( \angle B = 90^\circ \)
- \( \angle C = 45^\circ \)
- Triangle ABC is a right isosceles triangle.

---