In a triangle ABC, internal bisector of `angleABC` and external bisector of `angleACB` meet in D. Which one of the following is true ?

Options

A. `angleBDC=angleBAC`
B. `angleBDC=1/2angleABC`
C. `angleBDC=angleDBC`
D. None of these

Answer

To solve the problem, we need to analyze the triangle ABC and the angles formed by the internal bisector of angle ABC and the external bisector of angle ACB at point D. Let's break down the solution step by step.

### Step 1: Understand the Configuration
We have triangle ABC with:
- Internal bisector of angle ABC meeting at point D.
- External bisector of angle ACB also meeting at point D.

### Step 2: Define the Angles
Let:
- Angle A = ∠A
- Angle B = ∠B
- Angle C = ∠C
- Let the internal bisector of angle ABC create two angles at D: ∠ABD = ∠DBC = θ (since it is a bisector).
- The external bisector of angle ACB will create an external angle at point D, which we will denote as ∠ACD.

### Step 3: Apply the External Angle Theorem
According to the external angle theorem:
- The external angle ∠ACD = ∠A + ∠B (since it is equal to the sum of the two opposite interior angles).

### Step 4: Set Up the Equations
From the triangle ABC, we know:
- The sum of angles in triangle ABC is 180°:
\[ \angle A + \angle B + \angle C = 180° \]

From the external angle theorem, we can express ∠ACD:
\[ \angle ACD = \angle A + 2θ \]

### Step 5: Analyze Triangle BCD
In triangle BCD, we have:
- The angles are ∠BCD = ∠C, ∠BDC = θ, and ∠ACD = ∠A + 2θ.
- The sum of angles in triangle BCD is also 180°:
\[ \angle BDC + \angle C + \angle ACD = 180° \]
Substituting the known values:
\[ \angle BDC + \angle C + (\angle A + 2θ) = 180° \]

### Step 6: Substitute and Solve
From the previous equation of the sum of angles in triangle ABC:
\[ \angle C = 180° - \angle A - \angle B \]
Substituting this into the equation for triangle BCD gives:
\[ \angle BDC + (180° - \angle A - \angle B) + (\angle A + 2θ) = 180° \]
This simplifies to:
\[ \angle BDC + 180° - \angle B + 2θ = 180° \]
Thus:
\[ \angle BDC = \angle B - 2θ \]

### Step 7: Relate Angles
From the earlier setup, we know:
- Since θ = ∠ABD = ∠DBC, we can say:
\[ \angle BDC = \frac{1}{2} \angle ABC \]

### Conclusion
After analyzing the angles and applying the external angle theorem, we find that:
\[ \angle BDC = \frac{1}{2} \angle ABC \]

Thus, the correct option is:
**BDC is equal to half of ABC.**