Question

Prove that the four triangles got by joining the midpoints of a triangle are all congruent and that they are similar to the original triangle.

Answer 1

Given: D, E and F are the midpoints of the sides BC, CA and AB respectively.

∴ BD = DC = , AE = EC = , AF = FB = …(1)

Since line segment DE passes through the midpoints of the side BC and AC, DE is parallel to AB and DE = .

⇒ DE = AF = FB … (2)

In ΔCDE and ΔDBF:

DC = BD (From equation (1))

∠CDE = ∠DBF (Corresponding angles)

BF = DE (From equation (2))

As the side, angle, and side of one triangle is equal to the corresponding side, angle, and side of the other triangle, ΔCDE ≅ ΔDBF.

Similarly, ΔFBD, ΔDEF and ΔAFE are all congruent to each other.

Now, in ΔDCE and ΔBCA:

∠CED = ∠CAB (Corresponding angles)

∠CDE = ∠CBA (Corresponding angles)

∠ECD = ∠ACB (Common angles)

As all the angles of the triangle are equal to the angles of the other triangle, both the triangles are similar.

Similarly, ΔBDF, ΔFEA and ΔEFD are all similar to ΔBCA.

Thus, the four triangles got by joining the midpoints of the triangle are all congruent and they are similar to the original triangle.