Question

Prove that if a straight line is drawn parallel to one side of a triangle to intersect the other two sides in two distinct points, then the other two sides are divided by those points in the same ratio.

Hint:

When a line is drawn parallel to a side of a triangle, think "similar triangles first." From $\triangle ADE \sim \triangle ABC$, convert $AB/AD=AC/AE$ into $AD/DB=AE/EC$ by subtracting and inverting.

Answer 1

Let $\triangle ABC$ be a triangle and let a line through $D\in AB$ and $E\in AC$ be drawn such that $DE\parallel BC$. We must prove $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.

Step 1: Similar triangles.
Since $DE\parallel BC$, we have $\angle ADE=\angle ABC$ and $\angle AED=\angle ACB$, and $\angle A$ is common.
Therefore, $\triangle ADE \sim \triangle ABC$.

Step 2: Proportional corresponding sides.
From similarity, \[ \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}. \tag{1} \]

Step 3: Convert to the required ratio.
From the first two equal ratios in (1), \[ \frac{AB}{AD}=\frac{AC}{AE} \ \Rightarrow\ \frac{AB-AD}{AD}=\frac{AC-AE}{AE} \ \Rightarrow\ \frac{DB}{AD}=\frac{EC}{AE}. \] Taking reciprocals gives \[ \boxed{\ \frac{AD}{DB}=\frac{AE}{EC}\ }, \] which proves that the two sides are cut in the same ratio.