In `DeltaABC, angleC=2angle A` and AC=2BC. Then which of the following is/are True ?

Options

A. Angles A,B,C are in arithmetic progression
B. Angles A,C,B are in arithmetic progression
C. `Delta ABC` is a right angled isosceles triangle
D. `BC^(2)+CA^(2)+AB^(2)=8R^(2)`, where R is the circum-radius of `Delta ABC`

Explanation

To solve the problem, we need to analyze the given conditions in triangle \( ABC \):

1. \( \angle C = 2 \angle A \)
2. \( AC = 2 BC \)

### Step 1: Set Up the Angles
Let \( \angle A = A \), then \( \angle C = 2A \). By the angle sum property of triangles, we have:
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting for \( \angle C \):
\[ A + \angle B + 2A = 180^\circ \]
This simplifies to:
\[ 3A + \angle B = 180^\circ \]
Thus, we can express \( \angle B \) as:
\[ \angle B = 180^\circ - 3A \]

### Step 2: Apply the Law of Sines
Using the Law of Sines, we have:
\[ \frac{AC}{\sin B} = \frac{BC}{\sin A} = \frac{AB}{\sin C} \]
Given \( AC = 2BC \), we can write:
\[ \frac{2BC}{\sin B} = \frac{BC}{\sin A} \]
Dividing both sides by \( BC \) (assuming \( BC \neq 0 \)):
\[ \frac{2}{\sin B} = \frac{1}{\sin A} \]
Cross-multiplying gives:
\[ 2 \sin A = \sin B \]

### Step 3: Substitute for \( \sin B \)
From the expression for \( \angle B \):
\[ \sin B = \sin(180^\circ - 3A) = \sin 3A \]
Thus, we have:
\[ 2 \sin A = \sin 3A \]

### Step 4: Use the Sine Triple Angle Formula
Using the triple angle formula for sine:
\[ \sin 3A = 3 \sin A - 4 \sin^3 A \]
Substituting this into our equation gives:
\[ 2 \sin A = 3 \sin A - 4 \sin^3 A \]
Rearranging this leads to:
\[ 0 = \sin A - 4 \sin^3 A \]
Factoring out \( \sin A \):
\[ \sin A (1 - 4 \sin^2 A) = 0 \]

### Step 5: Solve for \( \sin A \)
This gives us two cases:
1. \( \sin A = 0 \) (not possible in a triangle)
2. \( 1 - 4 \sin^2 A = 0 \)

Solving \( 1 - 4 \sin^2 A = 0 \):
\[ 4 \sin^2 A = 1 \implies \sin^2 A = \frac{1}{4} \implies \sin A = \frac{1}{2} \]
Thus, \( A = 30^\circ \).

### Step 6: Find \( \angle C \) and \( \angle B \)
Using \( A = 30^\circ \):
\[ \angle C = 2A = 60^\circ \]
Now substituting back to find \( \angle B \):
\[ \angle B = 180^\circ - 3 \times 30^\circ = 90^\circ \]

### Conclusion
The angles of triangle \( ABC \) are:
- \( \angle A = 30^\circ \)
- \( \angle B = 90^\circ \)
- \( \angle C = 60^\circ \)