## Basic Proportionality Theorem:

**Theorem 1:** In a triangle a line drawn parallel to one side, to intersect the other two sides then it divides the two sides in the same ratio.

**Given:** A △ABC in which DE || BC and intersects AB and AC in D and E respectively.

**To prove:** AD/DB = AE/EC

**Construction:** Join BE and CD. Draw EF ⊥ AB and DG ⊥ AC.

**Proof:** In the figure, Area (△ADE) = (1/2) AD . EF

(Here AD is considered as the base of △ADE and EF is the altitude).

Area (△BDE) = (1/2) DB . EF

(Here DB is the base of △BDE and EF is the altitude).

∴ Area (△ADE)/Area (△BDE) = [(1/2) AD . EF]/[(1/2) DB . EF] = AD/DB ……….(i)

Again, we write area (△ADE) = (1/2) AE . DG

(Here AE is considered as base and DG as the altitude of △ADE)

Also, area (△CDE) = (1/2) EC . DG

(Here EC is considered as base of △CDE and DG is the altitude)

∴ Area (△ADE)/Area (△CDE) = [(1/2) AE . DG]/[ (1/2) EC . DG] = AE/EC ……….(ii)

Since △BDE and △CDE are on the same base DE and between the same parallel lines DE and BC,

∴ area (△BDE) = area (△CDE) ……….(iii)

From (i), (ii) and (iii) it follows that AD/DB = AE/EC

Hence, the theorem is proved.

Corollary: If in a △ABC, a line DE || BC, intersects AB in D and AC in E, then (i) AB/AD = AC/AE (ii) AB/DB = AC/EC.Proof: From Basic Proportionality theorem, we have(i) AD/DB = AE/EC ⇒ DB/AD = EC/AE ⇒ 1 + DB/AD = 1 + EC/AE [∵ Adding 1 to both sides] ⇒ (AD + DB)/AD = (AE + EC)/AE ⇒ AB/AD = AC/AE (ii) AD/DB = AE/EC ⇒ 1 + AD/DB = 1 + AE/EC [∵ Adding 1 to both sides] ⇒ (DB + AD)/DB = (EC + AE)/EC ⇒ AB/DB = AC/EC |

**Theorem 2:** (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.

**Given:** A △ABC and a line ‘*l*‘ intersecting AB in D and AC in E such that AD/DB = AE/EC.

**To prove:** *l* || BC.

**Proof:** If possible, we suppose that the line ‘*l*‘ is not parallel to BC. Therefore, there is some other line through D, which is parallel to BC. Let this line be DF. Since DF || BC, therefore, by Thales theorem, we get

AD/DB = AF/FC ………(i)

and AD/DB = AE/EC (given) ………(ii)

∴ (i) and (ii) imply that

AF/FC = AE/EC

⇒ 1 + AF/FC = 1 + AE/EC [∵ Adding 1 to both sides]

⇒ (FC + AF)/FC = (EC + AE)/EC

⇒ AC/FC = AC/EC ………(iii)

∵ (iii) is possible only if the points F and E coincides.

∴ DF is the line ‘*l*‘ itself. Hence *l* || BC.