Prove: If a line is drawn parallel to one side of a triangle to meet the other two sides, then it divides those sides in the same ratio.
Proved.
Step 1: Draw a triangle \(\triangle ABC\).
Mark a point \(D\) on side \(AB\) and a point \(E\) on side \(AC\). Draw line \(DE\) such that \(DE \parallel BC\).
Step 2: Because \(DE \parallel BC\), angle \(ADE\) is equal to angle \(ABC\) (corresponding angles), and angle \(AED\) is equal to angle \(ACB\) (corresponding angles).
Step 3: So, by the AA (Angle–Angle) similarity criterion, we can say:
\(\triangle ADE \sim \triangle ABC\).
Step 4: If two triangles are similar, then their sides are in proportion. Therefore,
\[ \dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}. \]
Step 5: From the first two parts of this equality, we get:
\[ \dfrac{AD}{AB} = \dfrac{AE}{AC}. \]
Step 6: Write \(AB = AD + DB\) and \(AC = AE + EC\). Substitute these values:
\[ \dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC}. \]
Step 7: Cross multiply:
\[ AD \times (AE + EC) = AE \times (AD + DB). \]
Step 8: Expand both sides:
\[ AD \times AE + AD \times EC = AE \times AD + AE \times DB. \]
Step 9: Cancel the common term \(AD \times AE\) from both sides:
\[ AD \times EC = AE \times DB. \]
Step 10: Divide both sides by \(DB \times EC\):
\[ \dfrac{AD}{DB} = \dfrac{AE}{EC}. \]
Final Result: The line \(DE\) divides the two sides \(AB\) and \(AC\) in the same ratio.