Question

If \(A, B, C\) are vertices of a triangle with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively, then find the position vector of the point \(D\) where the angle bisector from vertex \(A\) meets \(BC\).

Hint:

Recall the Angle Bisector Theorem: the bisector from a vertex divides the opposite side in the ratio of the two sides meeting at that vertex. Once that ratio is known, apply the section formula for internal division to write the position vector of the point.

Answer 1

Step 1: Understanding the Concept:
The internal angle bisector of a triangle divides the opposite side in the ratio of the sides containing the angle.
In \(\triangle ABC\), the angle bisector \(AD\) of \(\angle A\) divides the side \(BC\) in the ratio \(AB : AC\).

Step 2: Key Formula or Approach:
1. Length of side \(AB = c' = |\vec{b} - \vec{a}|\).
2. Length of side \(AC = b' = |\vec{c} - \vec{a}|\).
3. Internal Section Formula: If \(D\) divides \(BC\) in ratio \(m : n\), then \(\vec{d} = \frac{m\vec{c} + n\vec{b}}{m + n}\).

Step 3: Detailed Explanation:
The ratio in which \(D\) divides \(BC\) is \(BD : DC = AB : AC\).
Let \(m = AB = |\vec{b} - \vec{a}|\) and \(n = AC = |\vec{c} - \vec{a}|\).
Wait, the ratio is \(BD : DC = m : n\).
Applying the section formula for internal division:
\[ \vec{d} = \frac{BD \cdot \vec{c} + DC \cdot \vec{b}}{BD + DC} \] Since \(BD/DC = AB/AC\), we can write \(BD = k \cdot AB\) and \(DC = k \cdot AC\).
\[ \vec{d} = \frac{(AB)\vec{c} + (AC)\vec{b}}{AB + AC} \] Substituting the vector magnitudes:
\[ \vec{d} = \frac{|\vec{b} - \vec{a}| \vec{c} + |\vec{c} - \vec{a}| \vec{b}}{|\vec{b} - \vec{a}| + |\vec{c} - \vec{a}|} \] Rearranging for standard order:
\[ \vec{d} = \frac{|\vec{c} - \vec{a}| \vec{b} + |\vec{b} - \vec{a}| \vec{c}}{|\vec{c} - \vec{a}| + |\vec{b} - \vec{a}|} \]
Step 4: Final Answer:
The position vector of \(D\) is \(\vec{d} = \frac{|\vec{c} - \vec{a}| \vec{b} + |\vec{b} - \vec{a}| \vec{c}}{|\vec{c} - \vec{a}| + |\vec{b} - \vec{a}|}\).

Answer 2

Concept:

• Instead of directly quoting the Angle Bisector Theorem, the ratio $BD:DC = AB:AC$ can be derived from areas: triangles $ABD$ and $ACD$ share the same perpendicular height from $A$ to line $BC$, so their area ratio equals $BD:DC$.
• Because $AD$ bisects angle $A$, the angles $\angle BAD$ and $\angle CAD$ are equal, so writing the same two areas using the $\dfrac{1}{2}\times\text{side}\times\text{side}\times\sin(\text{angle})$ formula makes the sine terms cancel, leaving the ratio in terms of the side lengths alone.

Step 1: Compare the areas of triangle $ABD$ and triangle $ACD$ using their common height from $A$.
Since $B, D, C$ are collinear, both triangles share the same perpendicular height from $A$ to line $BC$.
$\dfrac{\text{Area}(ABD)}{\text{Area}(ACD)} = \dfrac{BD}{DC}$

Step 2: Compute the same two areas using the angle at $A$ instead.
$\text{Area}(ABD) = \dfrac{1}{2}\cdot AB \cdot AD \cdot \sin(\angle BAD)$
$\text{Area}(ACD) = \dfrac{1}{2}\cdot AC \cdot AD \cdot \sin(\angle CAD)$
Since $AD$ bisects angle $A$, $\angle BAD = \angle CAD$, so their sines are equal and $AD$ is common to both.
$\dfrac{\text{Area}(ABD)}{\text{Area}(ACD)} = \dfrac{AB}{AC}$

Step 3: Equate the two expressions for the same ratio.
$\dfrac{BD}{DC} = \dfrac{AB}{AC} = \dfrac{c}{b}$, where $c = AB = |\vec{b} - \vec{a}|$ and $b = AC = |\vec{c} - \vec{a}|$.

Step 4: Apply the section formula using this ratio.
$D$ divides $BC$ internally in the ratio $BD:DC = c:b$ (from $B$ towards $C$), so:
$\vec{d} = \dfrac{c\,\vec{c} + b\,\vec{b}}{c + b}$

Final Answer: $\vec{d} = \dfrac{b\,\vec{b} + c\,\vec{c}}{b + c}$, where $b = |\vec{c} - \vec{a}|$ and $c = |\vec{b} - \vec{a}|$