Let A(x₁ , y₁), B(x₂ , y₂) and C(x₃ , y₃) be the vertices of triangle.

Let D(3, 4), E(4, 6) and F(5, 7) be the midpoints of sides BC, CA and AB respectively.

By midpoint formula.

x = , y =

For midpoint D(3, 4) of side BC,

3 = , 4 =

∴ = 6 and = 8 …(1)

For midpoint E(4, 6) of side CA,

4 = , 6 =

∴ = 8 and = 12 …(2)

For midpoint F(5, 7) of side AB,

5 = , 7 =

∴ = 10 and = 14 …(3)

Adding 1,2 and 3, we get,

=6 + 8 + 10

And = 8 + 12 + 14

∴ 2(= 24 and 2() = 34

∴ = 12 and = 17

+ 6 = 12 and + 8 = 17 …from 1

∴ = 6 and = 9

Substituting above values in 3,

6 + and 9 + = 14

∴ = 4 and = 5

Similarly for equation 2,

6 + = 8 and 9 + = 12

∴ = 2 and = 3

Hence the vertices of triangle are A(6 , 9), B(4 ,5) and C(2 ,3)