1
AD = DB and DE ∥ BC ⇒ AE = EC
2
AD = DB and AE = EC ⇒ DE ∥ BC and DE = ½BC
FORMULA CHAPTER
Chapter 09 – Transversal and Mid-Point Theorems
WBBSEClass 09Mathematics2 formulas
Complete WBBSE Madhyamik Class 9 Transversal and Mid-Point Theorems formula list with notation, conditions and worked examples for exam preparation.
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Converse of the Mid-Point Theorem
AD = DB and DE ∥ BC ⇒ AE = EC
SymbolMeaningUnit
ABCGiven triangleNot applicable
DMidpoint of ABNot applicable
EPoint where the parallel line meets ACNot applicable
DELine segment drawn parallel to BClength units
BCSide parallel to DElength units
ACSide bisected at Elength units
Worked example
Question: In △ABC, D is the midpoint of AB. Through D, DE is drawn parallel to BC and meets AC at E. If AC = 14 cm, find AE and EC.
Calculation:
Step 1: D is the midpoint of AB, so AD = DB.
Step 2: DE ∥ BC. By the converse of the mid-point theorem, E is the midpoint of AC.
Step 3: AE = EC = ½AC = ½ × 14 = 7 cm.
Answer: AE = EC = 7 cm.
Mid-Point Theorem
AD = DB and AE = EC ⇒ DE ∥ BC and DE = ½BC
SymbolMeaningUnit
ABCGiven triangleNot applicable
DMidpoint of ABNot applicable
EMidpoint of ACNot applicable
DESegment joining the midpointslength units
BCThird side of the trianglelength units
Worked example
Question: In △ABC, D and E are the midpoints of AB and AC. If BC = 12 cm, find DE and state its relation with BC.
Calculation:
Step 1: Since D and E are midpoints, apply the mid-point theorem.
Step 2: DE ∥ BC.
Step 3: DE = ½BC = ½ × 12 = 6 cm.
Answer: DE = 6 cm and DE ∥ BC.