Parallelograms


   
 
Theorem 7
 
Parallel lines and Triangles
 
So far we have proved various theorems on parallelograms. Let us now apply these theorems to prove a few interesting and useful facts about a triangle.
 
Statement
 
"The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it".
 
 
Given:
 
In D ABC, AD=DB and AE=EC
 
To prove:
 
 
Construction:
 
Analysis for construction shows Analysis for construction: that you have to draw CF||BD Think how you can complete to meet DE produced at F. a parallelogram with DB and BC as consecutive sides. You will find the need to draw CF||DB.
 
Proof:
 
 
In triangles, ADE and CEF,
 
AE=EC (given)
 
 
 
 
AD=CF and DE=EF (corresponding parts of corresponding triangles)
 
But AD=DB (given)
 
DB=CF ----(i)
 
(AD is equal to both DB and CF)
 
In quadrilateral DBCF,
 
DB=CF and DB||CF
 
DBCF is a parallelogram. (by definition of parallelogram)
 
 
DF=BC
 
(opposite sides of a parallelogram are equal)
 
and DF||BC ----(ii)
 
But DE = EF (proved above)
 
And DF=DE+EF
 
 
=2 DE
 
and DF=BC (from (ii))
 
BC=2 DE
 
 
Hence the theorem is proved.
 
  
     
   
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