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

Explanation

GIVEN `A Delta ABC` in which `D,E,F` are the midpoints of `BC, CA and AB` respectively.
TO PROOVE `Delta AFE~ Delta ABC`.
`Delta FBD~ Delta ABC`,
`Delta EDC~ Delta ABC`
and `Delta ~ Delta ABC`.
PROOF We shall first show that `Delta AEF~ Delta ABC`.
Since, F and E are midpoint of AB and AC respeectively , so by the midpoint theorem, we have `FE||BC`.
`:. angle AEF=angle B` [ corresponding `angle`]
Now, in `Delta AFE and Delta ABC`, we have
`angle AEF = angle B [ correspoding `angle`]
and ` angle A= angle A` [ common]
`:. Delta AFE~ Delta ABC` [ by AA- similarity]
Similarly, `Delta FBD~ Delta ABC and Delta EDC~ Delta ABC`.
Now, we shall manner as above, we can prove that
`ED||AF and DF||EA`.
`:. AFDE` is a ||gm.
`:. angle EDF=angle A`[ opposite angels of a||gm]
Similarly, `BDEF` is a||gm
`:. angle DEF= angle B` [ opposite angels of a ||gm]
Thus, in `angle DEF and Delta A and Delta DEF= angle B`.
`:. Delta DEF~ DeltaABC` [ by AA-similarity]
Hence, the result followes.