In `DeltaABC, /_B = 2/_C`. D is a point on side BC such that AD bisects `/_BAC and AD = CD`. Find the value of `/_BAC`.

Options

A. `36^@`
B. `72^@`
C. `108^@`
D. `120^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Angles
Let the angle \( C \) be \( x \). According to the problem, angle \( B \) is twice angle \( C \), so we can write:
\[ B = 2x \]

### Step 2: Express Angle \( A \)
The sum of the angles in triangle \( ABC \) is \( 180^\circ \). Therefore, we can express angle \( A \) as:
\[ A = 180^\circ - B - C = 180^\circ - 2x - x = 180^\circ - 3x \]

### Step 3: Use the Angle Bisector Theorem
Point \( D \) is on side \( BC \) such that \( AD \) bisects angle \( BAC \). This means:
\[ \angle BAD = \angle CAD = \frac{A}{2} = \frac{180^\circ - 3x}{2} \]

### Step 4: Set Up the Relationship Between Angles
Since \( AD = CD \), triangles \( ACD \) and \( ADC \) are isosceles. Therefore, the angles opposite to equal sides are equal:
\[ \angle ACD = \angle CAD \]
Let \( \angle CAD = \angle ACD = \frac{180^\circ - 3x}{2} \).

### Step 5: Calculate the Angles in Triangle \( ACD \)
In triangle \( ACD \), the angles are:
- \( \angle CAD = \frac{180^\circ - 3x}{2} \)
- \( \angle ACD = \frac{180^\circ - 3x}{2} \)
- \( \angle ADC = 180^\circ - \angle CAD - \angle ACD = 180^\circ - 2 \cdot \frac{180^\circ - 3x}{2} = 3x \)

### Step 6: Apply the Angle Sum Property
Now, we have the angles in triangle \( ACD \):
\[ \angle A + \angle C + \angle D = 180^\circ \]
Substituting the angles we found:
\[ \frac{180^\circ - 3x}{2} + \frac{180^\circ - 3x}{2} + 3x = 180^\circ \]
This simplifies to:
\[ 180^\circ - 3x + 3x = 180^\circ \]
This equation holds true, confirming our angle relationships.

### Step 7: Solve for \( x \)
Now we can use the relationship established earlier:
\[ 5x = 180^\circ \]
Thus,
\[ x = \frac{180^\circ}{5} = 36^\circ \]

### Step 8: Find Angle \( A \)
Now we can find angle \( A \):
\[ A = 180^\circ - 3x = 180^\circ - 3 \cdot 36^\circ = 180^\circ - 108^\circ = 72^\circ \]

### Final Answer
The value of angle \( BAC \) is:
\[ \angle BAC = 72^\circ \]

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