Question

State and Prove BPT? With extra statement

Answer 1

In the figure alongside, if we consider DE is parallel to BC, then according to the theorem,

ADBD=AECE

Let’s not stop at the statement, we need to find a proof that its true. So shall we begin?

PROOF OF BPT

Given: In  ΔABC, DE is parallel to BC

Line DE intersects sides AB and PQ in points D and E, such that we get triangles A-D-E and A-E-C.

To Prove: ADBD=AECE

Construction: Join segments DC and BE

Proof: 

In ΔADE and ΔBDE,

A(ΔADE)A(ΔBDE)=ADBD                 (triangles with equal heights)

In ΔADE and ΔCDE,

A(ΔADE)A(ΔCDE)=AECE                  (triangles with equal heights)

Since ΔBDE and ΔCDE have a common base DE and have the same height we can say that,

A(ΔBDE)=A(ΔCDE)

Therefore,

A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)

Therefore,

ADBD=AECE

Hence Proved.

The BPT also has a converse which states, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Answer 2

Answer:

It states that if a line passing through two sides of triangle then if it is parallel to third side then it divides other two sides in same ratio.

proof is in the ATTACHMENT