Question

In ΔABC, D, E and F are midpoints of BC, CA and AB respectively. Prove that ΔFBD ∼ ΔDEF and ΔDEF ∼ ΔABC.

Answer 1

we know that,

The line joining the midpoints of two sides of a triangle is parallel to the 3rd side.

In ΔABC,

F and E are mid-points of AB and AC, respectively,

∴ FE || BC

So, FE || BD ...(1)

Similarly,

D and E are midpoints of BC and AC, respectively.

∴ DE || AB

So, DE || BF ....(2)

from (1) and (2)

FE || BD and DE || BF

Therefore, the opposite sides of the quadrilateral are parallel.

∴ DBEF is a parallelogram.

Similarity,

DCEF is a parallelogram.

Since, DBEF is a parallelogram, and in parallelogram, opposite angles are equal,

∴ ∠DFF = ∠ABC .....(3)

Since DCEF is a parallelogram, and in a parallelogram, opposite angles are equal.

∴ ∠DFF = ∠ACB .....(4)

In ΔDEF and ΔABC

∠DEF and ∠ACB

∠DEF = ∠ABC

∴ ΔDEF ∽ ΔABC ...[AA Similarity Criterion]

In ΔFBD and ΔDEF

∠FBD = ∠DEF

 ∠BDF = ∠EFD ...[∵ Alternate angles]

∴ ΔFBD ∽ ΔDEF ...[AA Similarity Criterion]