Question

converse of Thales theorem

Answer 1

____HEYA ____

Proof of the converse of "Thales Theorem":

First, let the line d intersect the sides AB and AC of △ABC at distinct points E and G, respectively, such that AEEB=AGGC. We now need to prove that EG∥BC.

Assume EG∦BC. Then there must be another line intersecting point E of side AB as well as some point, say F, of side AC that is parallel to BC. So, let EF∥BC.

By Thales Theorem, since EF∥BC, it follows that:
AEEB=AFFC(1)
But we are given
AEEB=AGGC(2)
Hence, from (1) and (2), it must follow that
AFFC=AGGC(3)
Adding "1" to both sides of equation (3) gives us:
AFFC+FCFC=AGGC+GCGC
which simplifies to AF+FCFC=AG+GCGC⟹ACFC=ACGC⟹FC=GC
But FC=GC is only possible when points F and G coincide with one another, i.e. if EF is the line d=EG itself.

But EF∥BC, and hence it cannot be the case that EG=EF∦BC.

Therefore, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side: i.e. the converse of Thales Theorem has been established.

Answer 2

Answer:

Theorem :- If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given :- A triangle ABC in which AD/DB=AE/EC. ---------(1)

To Prove :- DE||BC.

Construction :- Draw a line DF.

Proof :- Suppose DE is not parallel to BC.

Let DF||BC.

Then by Thales Theorem, we have

AD/BD = AF/FC.  --------(2)

FROM (1) & (2)

AE/EC = AF/FC

Adding 1 to both sides

AE/EC +1 = AF/FC +1

AE+EC/EC = AF+FC/FC

AC/EC = AC/FC

AC Cancelled out from both sides

EC = FC

This is not possible  as E and F are distinct points.

Thus are supposition is wrong.

DE || BC.

The diagram is below