**Theorem 6.1 Thale’s Theorem/ Basic Proportionality Theorem (BPT)**

* Statement*: ” If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points. The other two sides are divided in the same ratio.”

OR

Prove that in a triangle a line parallel to one side divides remaining two sides in the same ratio

**Given:**

In Δ ABC

DE // ( parallel to ) BC

**To Prove:**

AD/BD = AE/CE

**Construction:**

DF ⊥ ( Perpendicular ) on AC

EG ⊥ ( Perpendicular ) on AB

Join B to E and C to D

**Proof:**

According to figure:

In Δ ADE and Δ BDE

have common altitude (Perpendicular) = EG

∴ area Δ ADE /area Δ BDE = 1/2×b×h / 1/2×b×h

⇒ area Δ ADE /area Δ BDE = AD×EG / BD×EG

⇒ **area Δ ADE /area Δ BDE = AD / BD** ————-equation (i)

Similarly

In Δ ADE and Δ CED

Have common altitude(perpendicular) = DF

ar Δ ADE / ar Δ CED = 1/2×b×h / 1/2×b×h

⇒ ar Δ ADE / ar Δ CED = AE×DF / CE×DF

**⇒ ar Δ ADE / ar Δ CED = AE / CE ————-equation (ii)**

Δ BDE and Δ CED are lies on Common base (DE) and between same parallel lines

∴ ar Δ BDE = ar Δ CED

Dividing both side by area Δ ADE

⇒ ar Δ BDE/ar Δ ADE = ar Δ CED/ar Δ ADE

⇒ ar Δ ADE/ar Δ BDE = ar Δ ADE/ar Δ CED [ After reciprocal of both side] ————–equation(iii)

By comparing equation (i) and equation(ii) with equation (iii) We get

**AD/BD = AE/CE Hence proved**