Question

state converse of thasle theorem

Answer 1

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

Converse of "Thales Theorem":

First, let the line intersect the sides and of at distinct points E and G, respectively, such that We now need to prove that .

Assume . Then there must be another line intersecting point of side as well as some point, say , of side that is parallel to . So, let .

By Thales Theorem, since , it follows that:

But we are given

Hence, from (1) and (2), it must follow that

Adding "1" to both sides of equation (3) gives us:

which simplifies to

But is only possible when points and coincide with one another, i.e. if is the line itself.

But , and hence it cannot be the case that .

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