In triangle ABC; D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm, find the perimeter of the parallelogram BDEF.

By mid-point theorem,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side.

Given,

E is mid-point of AC and EF || AB.

∴ F is mid-point of BC (By converse of mid-point theorem).

Since, D and E are mid-points of sides AB and AC respectively.

∴ DE || BC and DE = $\dfrac{1}{2}BC$ (By mid-point theorem)

⇒ DE || BF and DE = BF (As F is mid-point of BC).

Given,

EF || BC

∴ EF || BD.

Since, E and F are mid-points of sides AC and BC respectively.

∴ EF = $\dfrac{1}{2}AB$ = BD. (By mid-point theorem)

Since, opposite sides of quadrilateral BDEF are parallel and equal.

∴ BDEF is a parallelogram.

From figure,

⇒ BD = $\dfrac{1}{2}AB = \dfrac{1}{2} \times 16$ = 8 cm,

⇒ BF = $\dfrac{1}{2}BC = \dfrac{1}{2} \times 18$ = 9 cm.

Perimeter of BDEF = BD + DE + EF + BF

= BD + BF + BD + BF (Since opposite sides of parallelogram are equal)

= 8 + 9 + 8 + 9

= 34 cm.

Hence, perimeter of parallelogram BDEF = 34 cm.