Long Answer Type
Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.
To Prove: Line segments AF and EC trisect the diagonal BD.
Proof: 
| Opposite sides of 
gm ABCD
AE 
FC .....(1)
AB = DC
| Opposite sides of 
gm ABCD
| Halves of equals are equal
AE = CF ....(2)
In view of (1) and (2),
AECF is a parallelogram
| A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of equal length
∴ EC || AF ...(3)
| Opposite sides of || gm AECF
In ∆DQC,
∵ F is the mid-point of DC
and FP || CQ | ∵ EC || AF
∴ P is the mid-point of DQ
| By converse of mid-point theorem
⇒ DP = PQ ...(4)
Similarly, in ∆BAP,
BQ = PQ ...(5)
From (4) and (5), we obtain
DP = PQ = BQ
⇒ Line segments AF and EC trisect the diagonal BD.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the mid-point of AC (ii) MD ⊥ AC (iii) 
Short Answer Type
Prove that MN = 
Long Answer Type
D, E, F are respectively the midpoints of the sides BC, CA and AB of a triangle ABC. Show that:
(i) BDEF is a parallelogram
(ii) DFEC is a parallelogram.
Short Answer Type
Long Answer Type
(AB + DC).
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