Onkar and write the relation between the lines joining the mid point of two sides with the third side of the triangle state the theorem to justify your answer why are the areas of the two fields is in the ratio of exactly 1: 3

When sides are the same then the triangles are congruent.
A1B1 = AB/2,  B1C1 = BC/2,  C1A1 = CA/2
where A1, B1 and C1 are the midpoints of the sides of Δ ABC.
1 ) A1B1 = C1B,  C1B1 = C1B1,  B1C1 = A1B
Three sides are equal ( SSS ) so ΔA1B1C1 = ΔBA1C1
2 ) A1C1=AB1, B1C1=B1C1, A1B1=AC1
SSS, ΔA1B1C1 = ΔAB1C1
3 ) B1C1 = CA1, B1A1=B1A1, A1C1= CB1
SSS, ΔA1B1C1 = ΔCA1B1
Therefore: ΔABC = 4 * ΔA1B1C1.
Area of ΔA1B1C1  : Area of the rest of ΔABC = 1 : 3
Picture is in the attachment.

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