- A5
- B8
- C3
- D4
Step-by-Step Solution
By BPT, AD/DB = AE/EC → . Cross-multiply: → → → .
Answer: D) 4
In ΔABC, DE ∥ BC. AD = , DB = , AE = and EC = . Find the value of .
Class 10Triangles (Similarity)Geometry — Similarity
Key idea: Basic Proportionality Theorem (BPT / Thales)
**Theorem**: If a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, then it divides them in the same ratio.
**Setup**: ΔABC. Line DE ∥ BC, with D on AB and E on AC.
**Conclusion**: AD/DB = AE/EC.
**Quick proof sketch**:
Draw DE ∥ BC. Triangles ADE and ABC have ∠A common and ∠ADE = ∠ABC (corresponding angles, DE ∥ BC). So ΔADE ~ ΔABC by AA.
Thus AD/AB = AE/AC.
From AD/AB = AE/AC, you can derive AD/DB = AE/EC by simple algebra.
**Converse**: if AD/DB = AE/EC, then DE ∥ BC.
Useful for proving lines parallel and finding lengths in triangles split by parallels.
**Setup**: ΔABC. Line DE ∥ BC, with D on AB and E on AC.
**Conclusion**: AD/DB = AE/EC.
**Quick proof sketch**:
Draw DE ∥ BC. Triangles ADE and ABC have ∠A common and ∠ADE = ∠ABC (corresponding angles, DE ∥ BC). So ΔADE ~ ΔABC by AA.
Thus AD/AB = AE/AC.
From AD/AB = AE/AC, you can derive AD/DB = AE/EC by simple algebra.
**Converse**: if AD/DB = AE/EC, then DE ∥ BC.
Useful for proving lines parallel and finding lengths in triangles split by parallels.
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