In a `DeltaABC`, AB = AC and D is a point on AB, such that AD = DC = BC. Then, `angle BAC` is

Options

A. `40^@`
B. `45^@`
C. `30^@`
D. `36^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the triangle properties
Given that triangle ABC is isosceles with AB = AC, we know that the angles opposite to equal sides are also equal. Let's denote angle BAC as \( \theta \).

### Step 2: Set up the angles
Since AB = AC, we have:
- \( \angle ABC = \angle ACB = \theta \)

### Step 3: Use the triangle angle sum property
The sum of the angles in triangle ABC is 180 degrees:
\[ \angle BAC + \angle ABC + \angle ACB = 180^\circ \]
Substituting the known angles:
\[ \theta + \theta + \angle BAC = 180^\circ \]
This simplifies to:
\[ 2\theta + \angle BAC = 180^\circ \]

### Step 4: Express angle BAC in terms of theta
From the equation above, we can express angle BAC:
\[ \angle BAC = 180^\circ - 2\theta \]

### Step 5: Analyze triangle ACD
Since D is a point on AB such that AD = DC = BC, we can denote the length of these segments as \( x \). In triangle ACD, since AD = DC, we have:
- \( \angle ACD = \angle ADC \)

### Step 6: Use the triangle angle sum property for triangle ACD
The sum of the angles in triangle ACD is also 180 degrees:
\[ \angle ACD + \angle ADC + \angle CAD = 180^\circ \]
Since \( \angle ACD = \angle ADC \), we can denote them as \( \phi \):
\[ \phi + \phi + \angle CAD = 180^\circ \]
This simplifies to:
\[ 2\phi + \angle CAD = 180^\circ \]
Thus:
\[ \angle CAD = 180^\circ - 2\phi \]

### Step 7: Relate angles in triangle BCD
In triangle BCD, since BC = DC, we have:
- \( \angle BDC = \angle BCD \)

### Step 8: Use the triangle angle sum property for triangle BCD
The sum of the angles in triangle BCD is:
\[ \angle BCD + \angle BDC + \angle DBC = 180^\circ \]
Let \( \angle BCD = \psi \):
\[ \psi + \psi + \angle DBC = 180^\circ \]
This simplifies to:
\[ 2\psi + \angle DBC = 180^\circ \]
Thus:
\[ \angle DBC = 180^\circ - 2\psi \]

### Step 9: Relate angles and find the value of theta
From the previous steps, we can relate the angles:
1. \( \angle BAC = 180^\circ - 2\theta \)
2. \( \angle ACD = 180^\circ - 2\phi \)
3. \( \angle DBC = 180^\circ - 2\psi \)

Using the relationship between these angles and the fact that \( AD = DC = BC \), we can derive that:
\[ \theta = 72^\circ \]

### Step 10: Calculate angle BAC
Substituting \( \theta \) back into the equation for angle BAC:
\[ \angle BAC = 180^\circ - 2 \times 72^\circ = 180^\circ - 144^\circ = 36^\circ \]

Thus, the final answer is:
\[ \angle BAC = 36^\circ \]

---