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Question

Prove that the line segment joining the mid points of the sides of a triangle form four triangles, each of which is similar to the original triangle.

Solution
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Given that:-
ABC in whichD, E, F are the mid points of sides AB, BC and CA respectively.
To prove:- each of the triangles are similar to the original triangle, i.e.,
ADFABC
BDEABC
CEFABC
Proof:-
Consider the ADF and ABC
Since D and F are the mid points of AB and AC respectively.
DFBC
AFD=B(Corresponding angles are equal)
Now, in ADF and ABC, we have
ADF=B(Corresponding angles)
A=A(Common)
By AA similar conditions,
ADFABC
Similarly, we have
BDEABC
CEFABC
EFAB
EFAD..........(1)
And, DEAC
DEAF..........(2)
From eqn(1)&(2), we have
ADEF is a parallelogram.
Similarly, BDFE is a parallelogram.
Now, in ABC and DEF
A=FED(Opposite angles of parallelogram)
B=DFE(Opposite angles of parallelogram)
Therefore, by AA similar condition
ABCDEF
Hence proved that each of the triangles are similar to the original triangle.

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