In ΔABC, the sides AB, AC are produced and the bisectors of exterior angles of ∠ABC and ∠ACB intersect at D. If ∠BAC = 50°, then ∠BDC is equal to
65°
Understanding the Triangle Geometry Problem
The problem asks us to find the measure of an angle formed by the intersection of the exterior angle bisectors of a triangle. We are given a triangle \( \Delta \text{ABC} \), where the sides AB and AC are extended. The bisectors of the exterior angles at vertices B and C meet at a point D. We are given that the interior angle \( \angle \text{BAC} = 50^\circ \), and we need to find the measure of \( \angle \text{BDC} \).
Key Concept: Exterior Angle Bisectors
There is a specific theorem in triangle geometry that relates the angle formed by the intersection of two exterior angle bisectors to the third interior angle of the triangle.
The theorem states that the angle formed by the intersection of the bisectors of two exterior angles of a triangle is equal to \( 90^\circ \) minus half of the third interior angle.
In \( \Delta \text{ABC} \), if the bisectors of the exterior angles at B and C intersect at D, then the angle formed at D, which is \( \angle \text{BDC} \), can be calculated using the formula:
\[ \angle \text{BDC} = 90^\circ - \frac{1}{2} \angle \text{BAC} \]Applying the Formula to Find ∠BDC
In this problem, we are given \( \angle \text{BAC} = 50^\circ \). We can directly use the formula mentioned above to find \( \angle \text{BDC} \).
Substitute the given value of \( \angle \text{BAC} \) into the formula:
\[ \angle \text{BDC} = 90^\circ - \frac{1}{2} \times 50^\circ \]First, calculate half of \( \angle \text{BAC} \):
\[ \frac{1}{2} \times 50^\circ = 25^\circ \]Now, subtract this value from \( 90^\circ \):
\[ \angle \text{BDC} = 90^\circ - 25^\circ \] \[ \angle \text{BDC} = 65^\circ \]So, the measure of \( \angle \text{BDC} \) is \( 65^\circ \).
Step-by-Step Calculation
Let's break down the calculation:
- Identify the given interior angle: \( \angle \text{BAC} = 50^\circ \).
- Identify the angle to be found: \( \angle \text{BDC} \), which is formed by the exterior angle bisectors at B and C.
- Recall the formula for the angle formed by two exterior angle bisectors: \( \angle \text{BDC} = 90^\circ - \frac{1}{2} \angle \text{BAC} \).
- Substitute the value of \( \angle \text{BAC} \): \( \angle \text{BDC} = 90^\circ - \frac{1}{2} (50^\circ) \).
- Calculate half of the angle: \( \frac{1}{2} \times 50^\circ = 25^\circ \).
- Perform the subtraction: \( \angle \text{BDC} = 90^\circ - 25^\circ \).
- Final result: \( \angle \text{BDC} = 65^\circ \).
Conclusion on Angle BDC
Using the property of the angle formed by the intersection of exterior angle bisectors, we found that \( \angle \text{BDC} \) is equal to \( 65^\circ \) when \( \angle \text{BAC} = 50^\circ \).
| Given Information | Required Angle | Formula Used | Result |
|---|---|---|---|
| \( \Delta \text{ABC} \) | \( \angle \text{BDC} \) | \( \angle \text{BDC} = 90^\circ - \frac{1}{2} \angle \text{BAC} \) | \( 65^\circ \) |
| \( \angle \text{BAC} = 50^\circ \) | |||
| Exterior angle bisectors at B and C meet at D |
Revision Table: Triangle Angles
| Concept | Description | Formula/Property |
|---|---|---|
| Sum of Interior Angles | Sum of angles inside any triangle is \( 180^\circ \). | \( \angle \text{A} + \angle \text{B} + \angle \text{C} = 180^\circ \) |
| Exterior Angle | An angle formed by extending one side of the triangle. It is equal to the sum of the two opposite interior angles. | Exterior angle at C = \( \angle \text{A} + \angle \text{B} \) |
| Interior Angle Bisector Meeting Point (Incenter) | Intersection of interior angle bisectors. Distance to all sides is equal. | Angle formed by two interior bisectors (e.g., \( \angle \text{BIC} \)) = \( 90^\circ + \frac{1}{2} \angle \text{BAC} \) |
| Exterior Angle Bisector Meeting Point (Excenter) | Intersection of two exterior angle bisectors and the bisector of the third interior angle. | Angle formed by two exterior bisectors (e.g., \( \angle \text{BDC} \)) = \( 90^\circ - \frac{1}{2} \angle \text{BAC} \) |
Additional Information: Exploring Exterior Angle Bisectors
When we extend the sides AB and AC of \( \Delta \text{ABC} \), we create exterior angles at vertices B and C. Let's denote the exterior angle at B as \( \angle \text{CBE'} \) (where E' is a point on the extension of AB) and the exterior angle at C as \( \angle \text{BCF'} \) (where F' is a point on the extension of AC).
The bisector of \( \angle \text{CBE'} \) is a ray that divides this exterior angle into two equal parts. Similarly, the bisector of \( \angle \text{BCF'} \) divides that exterior angle into two equal parts. The point D is where these two bisectors meet.
The relationship \( \angle \text{BDC} = 90^\circ - \frac{1}{2} \angle \text{BAC} \) is a standard result derived using the properties of angles on a straight line and the angle sum property of a triangle.
Let the bisector of the exterior angle at B be BD, and the bisector of the exterior angle at C be CD.
The exterior angle at B is \( 180^\circ - \angle \text{ABC} \). Since BD bisects it, \( \angle \text{DBC} = \frac{1}{2} (180^\circ - \angle \text{ABC}) = 90^\circ - \frac{1}{2} \angle \text{ABC} \).
Similarly, the exterior angle at C is \( 180^\circ - \angle \text{ACB} \). Since CD bisects it, \( \angle \text{DCB} = \frac{1}{2} (180^\circ - \angle \text{ACB}) = 90^\circ - \frac{1}{2} \angle \text{ACB} \).
Now, consider \( \Delta \text{BDC} \). The sum of angles in \( \Delta \text{BDC} \) is \( 180^\circ \):
\[ \angle \text{BDC} + \angle \text{DBC} + \angle \text{DCB} = 180^\circ \]Substitute the expressions for \( \angle \text{DBC} \) and \( \angle \text{DCB} \):
\[ \angle \text{BDC} + (90^\circ - \frac{1}{2} \angle \text{ABC}) + (90^\circ - \frac{1}{2} \angle \text{ACB}) = 180^\circ \] \[ \angle \text{BDC} + 180^\circ - \frac{1}{2} (\angle \text{ABC} + \angle \text{ACB}) = 180^\circ \]Subtract \( 180^\circ \) from both sides:
\[ \angle \text{BDC} - \frac{1}{2} (\angle \text{ABC} + \angle \text{ACB}) = 0 \] \[ \angle \text{BDC} = \frac{1}{2} (\angle \text{ABC} + \angle \text{ACB}) \]We know that in \( \Delta \text{ABC} \), \( \angle \text{BAC} + \angle \text{ABC} + \angle \text{ACB} = 180^\circ \). So, \( \angle \text{ABC} + \angle \text{ACB} = 180^\circ - \angle \text{BAC} \).
Substitute this into the equation for \( \angle \text{BDC} \):
\[ \angle \text{BDC} = \frac{1}{2} (180^\circ - \angle \text{BAC}) \] \[ \angle \text{BDC} = \frac{1}{2} \times 180^\circ - \frac{1}{2} \angle \text{BAC} \] \[ \angle \text{BDC} = 90^\circ - \frac{1}{2} \angle \text{BAC} \]This derivation confirms the formula used to solve the problem.
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