In triangle ABC, AB = 30 cm, AC = 45 cm and BC = 50 cm. The bisector of angle A meets BC at D. (a) Find BD and CD. (b) Through D, DE is drawn parallel to AB with E on AC, and DF parallel to AC with F on AB. Find DE and DF.

Step-by-step solution

1. 1

   Apply the angle bisector theorem. An internal bisector from $A$ divides the opposite side in the ratio of the two adjacent sides:

   $\frac{BD}{CD}=\frac{AB}{AC}=\frac{30}{45}=\frac{2}{3}$

   Note it is not the ratio of the other two sides, and not a midpoint — the bisector lands nearer the shorter side.

2. 2

   Turn the ratio into lengths. Write $BD=2x$ and $CD=3x$. Since $D$ lies on segment $BC$, the two pieces add to the whole:

   $2x+3x=50\quad\Longrightarrow\quad 5x=50\quad\Longrightarrow\quad x=10$

   so

   $BD=20\ \text{cm},\qquad CD=30\ \text{cm}$

   Check: $20+30=50$ ✓ and $20:30=2:3$ ✓.

3. 3

   Find DE using similar triangles. Because $DE\parallel AB$, the angles at $D$ and $E$ in $\triangle CDE$ match those at $B$ and $A$ in $\triangle CBA$ (corresponding angles), so $\triangle CDE\sim\triangle CBA$. Matching $DE$ with $AB$ and $CD$ with $CB$:

   $\frac{DE}{AB}=\frac{CD}{CB}\quad\Longrightarrow\quad \frac{DE}{30}=\frac{30}{50}\quad\Longrightarrow\quad DE=18\ \text{cm}$

   The key is pairing $DE$ with $AB$ (both opposite the angle at $C$'s vertex counterpart), not with $AC$.

4. 4

   Find DF the same way, from the other vertex. Because $DF\parallel AC$, we get $\triangle BDF\sim\triangle BCA$, so

   $\frac{DF}{AC}=\frac{BD}{BC}\quad\Longrightarrow\quad \frac{DF}{45}=\frac{20}{50}\quad\Longrightarrow\quad DF=18\ \text{cm}$

5. 5

   Explain why DE and DF come out equal. Both are $18$ cm, and that is no coincidence. With $DE\parallel AB$ and $DF\parallel AC$, the quadrilateral $AFDE$ is a parallelogram, so $AF=DE$ and $AE=DF$. But $AD$ bisects angle $A$ and is also a diagonal of that parallelogram, which forces the two adjacent sides to be equal — the parallelogram is a rhombus, so $DE=DF$. Algebraically, $AB\cdot\frac{CD}{BC}=AC\cdot\frac{BD}{BC}$ holds exactly because $\frac{BD}{CD}=\frac{AB}{AC}$.

6. 6

   Verify with coordinates. Placing $A=(0,0)$, $B=(30,0)$ and $C$ at the point with $AC=45$, $BC=50$, then locating $D$ at $\tfrac{20}{50}$ of the way from $B$ to $C$: the computed distances are $BD=20.000$, $CD=30.000$, the line $AD$ makes equal angles with $AB$ and $AC$, and the two parallels give $DE=18.000$ and $DF=18.000$ ✓.