Basic Proportionality A3F178

1. The problem asks us to prove the Basic Proportionality Theorem (BPT), also known as Thales' theorem. 2. The Basic Proportionality Theorem states: If a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. 3. Consider triangle $ABC$ with a line $DE$ parallel to side $BC$, intersecting $AB$ at $D$ and $AC$ at $E$. 4. We need to prove that $$\frac{AD}{DB} = \frac{AE}{EC}$$. 5. Since $DE \parallel BC$, angles $ADE$ and $ABC$ are corresponding angles and thus equal. 6. Similarly, angles $AED$ and $ACB$ are corresponding angles and equal. 7. Triangles $ADE$ and $ABC$ have two pairs of equal angles, so they are similar by the AA similarity criterion. 8. From similarity, corresponding sides are proportional, so $$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$$. 9. From the first two ratios, $$\frac{AD}{AB} = \frac{AE}{AC}$$. 10. Express $AB$ as $AD + DB$ and $AC$ as $AE + EC$. 11. Substitute to get $$\frac{AD}{AD + DB} = \frac{AE}{AE + EC}$$. 12. Cross-multiply: $$AD \times (AE + EC) = AE \times (AD + DB)$$. 13. Expand: $$AD \times AE + AD \times EC = AE \times AD + AE \times DB$$. 14. Cancel $AD \times AE$ from both sides: $$AD \times EC = AE \times DB$$. 15. Rearranged, $$\frac{AD}{DB} = \frac{AE}{EC}$$, which proves the theorem. Final answer: The Basic Proportionality Theorem is proved as $$\frac{AD}{DB} = \frac{AE}{EC}$$ when $DE \parallel BC$ in triangle $ABC$.