Question

If a line intersects the sides AB and AC at D and E respectively in a triangle ABC and it is parallel to the side BC, then prove that \[ \frac{AD}{AB} = \frac{AE}{AC} \]

Hint:

When a line parallel to one side of a triangle intersects the other two sides, it divides them proportionally (Basic Proportionality Theorem).

Answer 1

Step 1: Given.
In $\triangle ABC$, DE is parallel to BC. Step 2: By Basic Proportionality Theorem (Thales Theorem).
If a line is drawn parallel to one side of a triangle to intersect the other two sides, then it divides the two sides in the same ratio. Step 3: Apply the theorem.
Since $DE \parallel BC$, \[ \frac{AD}{DB} = \frac{AE}{EC} \]
Step 4: Simplify ratios.
Add 1 to both sides: \[ \frac{AD + DB}{DB} = \frac{AE + EC}{EC} \] \[ \frac{AB}{DB} = \frac{AC}{EC} \] By taking reciprocals, \[ \frac{AD}{AB} = \frac{AE}{AC} \]
Step 5: Conclusion.
Hence proved that $\dfrac{AD}{AB} = \dfrac{AE}{AC}$.