Converse of Basic Proportionality Theorem
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Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.|
If AD AE ---- = ------ then DE || BC DB EC |
Given : A Δ ABC and a line intersecting AB in D and AC in E,
such that AD / DB = AE / EC.
Prove that : DE || BC
Let DE is not parallel to BC. Then there must be another line that is parallel to BC.
Let DF || BC.
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| 1) DF || BC | 1) By assumption |
| 2) AD / DB = AF / FC | 2) By Basic Proportionality theorem |
| 3) AD / DB = AE /EC | 3) Given |
| 4) AF / FC = AE / EC | 4) By transitivity (from 2 and 3) |
| 5) (AF/FC) + 1 = (AE/EC) + 1 | 5) Adding 1 to both side |
| 6) (AF + FC )/FC = (AE + EC)/EC | 6) By simplifying |
| 7) AC /FC = AC / EC | 7) AC = AF + FC and AC = AE + EC |
| 8) FC = EC | 8) As the numerator are same so denominators are equal |
This is possible when F and E are same. So DF is the line DE itself.
∴ DF || BC
Examples
1) D and E are respectively the points on the sides AB and AC of a ΔABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, prove that DE || BC.
Solution :
AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm
∴ BD = AB – AD = 5.6 – 1.4 = 4.2 cm
And EC = AC – AE = 7.2 – 1.8 = 5.4 cm
Now, AD / DB =1.4 / 4.2 = 1/3
And AE / EC = 1.8 / 5.4 = 1/3
⇒ AD / DB = AE / EC
DE || BC ( by converse of basic proportionality theorem)
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2) State whether PQ || EF.DP / PE = 3.9 / 3 = 13/10
DQ / QF = 3.6 /2.4 = 3/2
So, DP / PE ≠ DQ / QF
⇒ PQ ∦ EF ( PQ is not parallel to EF)
Similarity in Triangles
• Similarity in Geometry
• Properties of similar triangles
• Basic Proportionality Theorem(Thales theorem)
• Converse of Basic Proportionality Theorem
• Interior Angle Bisector Theorem
• Exterior Angle Bisector Theorem
• Proofs on Basic Proportionality
• Criteria of Similarity of Triangles
• Geometric Mean of Similar Triangles
• Areas of Two Similar Triangles
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