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: Â 
         | 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 Â Â Â ...(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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