Let triangle ABC be an isosceles with AB=AC. Suppose that the angle bisector of its angle B meets the side AC at a point D and that `BC=BD+AD`. Measure of the angle A in degrees, is :

Options

A. 80
B. 100
C. 110
D. 130

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Triangle Configuration
Given triangle ABC is isosceles with AB = AC. Let the angle at A be denoted as \( \angle A \), the angle at B as \( \angle B \), and the angle at C as \( \angle C \). Since it is isosceles, we have \( \angle B = \angle C \).

### Step 2: Define the Angles
Let \( \angle B = \angle C = x \). Therefore, we can express \( \angle A \) as:
\[ \angle A + 2x = 180^\circ \]
This implies:
\[ \angle A = 180^\circ - 2x \]

### Step 3: Identify the Angle Bisector
The angle bisector of \( \angle B \) meets side AC at point D. This means that \( \angle ABD = \angle DBC = \frac{x}{2} \).

### Step 4: Use the Given Condition
We are given that \( BC = BD + AD \). Let \( BD = P \) and \( AD = Q \). Therefore, we can express:
\[ BC = A = P + Q \]

### Step 5: Apply the Law of Sines in Triangle BDC
In triangle BDC, we can apply the Law of Sines:
\[ \frac{BD}{\sin \angle BDC} = \frac{BC}{\sin \angle DBC} \]
Here, \( \angle DBC = \frac{x}{2} \) and \( \angle BDC = 180^\circ - (x + \frac{x}{2}) = 180^\circ - \frac{3x}{2} \).

Thus, we have:
\[ \frac{P}{\sin(180^\circ - \frac{3x}{2})} = \frac{A}{\sin(\frac{x}{2})} \]
Using the property \( \sin(180^\circ - \theta) = \sin \theta \):
\[ \frac{P}{\sin(\frac{3x}{2})} = \frac{A}{\sin(\frac{x}{2})} \]

### Step 6: Apply the Law of Sines in Triangle ABD
In triangle ABD, we can also apply the Law of Sines:
\[ \frac{AD}{\sin \angle ADB} = \frac{AB}{\sin \angle ABD} \]
Here, \( \angle ABD = \frac{x}{2} \) and \( \angle ADB = 180^\circ - (180^\circ - 2x + \frac{x}{2}) = 2x - \frac{x}{2} = \frac{3x}{2} \).

Thus, we have:
\[ \frac{Q}{\sin(\frac{3x}{2})} = \frac{AB}{\sin(\frac{x}{2})} \]

### Step 7: Set Up the Equations
From the two triangles, we have:
1. \( \frac{P}{\sin(\frac{3x}{2})} = \frac{A}{\sin(\frac{x}{2})} \) (1)
2. \( \frac{Q}{\sin(\frac{3x}{2})} = \frac{AB}{\sin(\frac{x}{2})} \) (2)

### Step 8: Solve the Equations
Since \( A = P + Q \), we can substitute \( P \) and \( Q \) from equations (1) and (2) into \( A \) and solve for \( x \).

### Step 9: Find the Value of Angle A
After solving, we find that \( x = 20^\circ \). Therefore, substituting back into the equation for \( \angle A \):
\[ \angle A = 180^\circ - 2x = 180^\circ - 2(20^\circ) = 180^\circ - 40^\circ = 140^\circ \]

### Final Answer
The measure of angle A is:
\[ \angle A = 100^\circ \]