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The Mid Point Theorem

Theorem 6 
In a triangle, the line segment joining the mid points of any two sides is parallel to the third side.
Let us look at the different types of quadrilaterals drawn below.

 

Given: In ΔABC, D is the mid-point of AB and E is the mid-point of AC.
To prove: DE = BC and DE
çç BC
Construction: Produce line-segment DE to F such that DE = EF and join FC.

 

Proof:
In
ΔADE and ΔCFE 
AE = EC       ...(given)
DE = EF      ...(by construction)

AED = CEF        ...(Vertically opposite s)
...
ΔADE ΔCFE
... AD = CF     ...(i)
and
ADE = CFE  ...(CPCT)
Since  AD = DB  ...(D is the mid-point)
and    AD = CF    ...(proved above)
... DB = FC        ...(ii)
Since
ADE = CFE
... AD
ïï CF ...   AB ïï CF  Þ DB ïï CF  ...(iii)
Now in quadrilateral BCFD,
DB = FC and DB
ïï FC
... BCFD is a parallelogram
BC = DF = DE + EF = 2 DE

...  DE = BC
Also,BC
ïï DF  ...(opposite sides of ||gm)
... BC ïï DE  Þ   DE ïï BC

 

Example :

In the figure M, N and O are the mid-points of sides PR, PQ and RQ of ΔPQR and PQ = QR = PR. Show that ΔMON is an equilateral triangle.

Solution :


Since M and N are the mid-points of PR and PQ of ΔPQR.
...  MN = RQ and MN
ïï RQ...(i)
Similarly, we can get
MO = PQ and MO
ïï PQ...(ii)
NO = PR and NO
ïï PR  ...(iii)
But PQ = QR = PR...(given)

...  PQ = QR = PR
...  MO = MN = NO
Hence,
ΔMNO is an equilateral triangle.
   

 




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