prove thales theorem converse
Answers
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.
If AD AE
---- = ------ then DE || BC
DB EC
Given : A Δ ABC and a line intersecting AB in D and AC in E,
such that AD / DB = AE / EC.
Prove that : DE || BC

Let DE is not parallel to BC. Then there must be another line that is parallel to BC.
Let DF || BC.
StatementsReasons1) DF || BC1) By assumption2) AD / DB = AF / FC2) By Basic Proportionality theorem3) AD / DB = AE /EC3) Given4) AF / FC = AE / EC4) By transitivity (from 2 and 3)5) (AF/FC) + 1 = (AE/EC) + 15) Adding 1 to both side6) (AF + FC )/FC = (AE + EC)/EC6) By simplifying7) AC /FC = AC / EC7) AC = AF + FC and AC = AE + EC8) FC = EC8) As the numerator are same so denominators are equal
This is possible when F and E are same. So DF is the line DE itself.
∴ DF || BC
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First, let the line dd intersect the sides AB and AC of △ABC at distinct points E and G, respectively, such that AE/EB=AG/GC.AE/EB=AG/GC. We now need to prove that EG∥BC
Assume EG∦BC Then there must be another line intersecting point EE of side ABAB as well as some point, say FF, of side ACAC that is parallel to BCBC. So, let EF∥BC
By Thales Theorem, since EF∥BC, it follows that:
AE/EB=AF/FC(1)
But we are given
AE/EB=AG/GC(2)
Hence, from (1) and (2), it must follow that
AF/FC=AG/GC(3)
Adding "1" to both sides of equation (3) gives us:
AF/FC+FC/FC=AG/GC+GC/GC
which simplifies to AF+FC/FC=AG+GC/GC
⟹AC/FC=AC/GC
⟹FC=GC
But FC=GC is only possible when points FF and GG 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.
