Mid Point Theorem


 
 
Concept Explanation
 

Mid Point Theorem

MID POINT THEOREM: 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 A large Delta ABC in which D and E are the mid-points of sides AB and AC respectively. DE is joined.

TO PROVE large DEparallel BC;and;DE=frac{1}{2}BC

CONSTRUCTION  Produce the line segment DE to F, such that DE = EF. Join FC.

PROOF  In large Delta sAED;and;CEF, we have

               AE = CE                              [large because E is the mid-point of AC]

          large angle AED=angle CEF    [ Vertically opposite angles]

and,        DE = EF                                 [By construction]

So, large Delta AEDcong Delta CEF    [By SAS criterion of congruence]

     large Rightarrow    AD = CF                                          [C.P.C.T.]                          .....(i)

and,   large angle ADE=angle CFE             [C.P.C.T.]                            .....(ii)

As it is given that D is the mid-point of AB

large Rightarrow ;;;DB=AD

large Rightarrow ;;;DB=CF         [From (i) we know that AD = CF]                 .....(iii)

From Eq 2 We know that Now, large angle ADE=angle CFE

But they are alternate interior angle when AB and CF are straight lines and DF is the transversal

As they are equal so AB || FC , or BD || CF   [ When whole are parallel the parts are parallel]

Hence BD = CF and BD || CF

Rightarrow BCFD is a IIgm  [ As one opposite pair of sides is parallel and equal]

large Rightarrow DF || BC   and DF = BC               [ large because  Opposite sides of a large parallel gm are equal and parallel]

But, E is the midpoint of DF   [ By construction DE = EF ]

large therefore     large DEparallel BC;and;DE=frac{1}{2}BC  Hence proved.

Illustration:ABCD is a rhombus, EABF is a straight line such that EA = AB = BF Prove that ED and CF when produced meet at right angle.

Proof:  ABCD is a rhombus and diagonals of a rhombus bisect at right angle

therefore angle AOB = 90^0

In Delta EDB

A is the mid point of EB   [ Given EA = AB]

O is the mid point of BD [ Diagonals bisect]

Now as O and A are mid points of BD and EB . By mid point theorem we can say that ED || AORightarrow EG || AC

Similarly  In Delta ACF

 B is the mid point of AF   [ Given AB = BF]

O is the mid point of AC [ Diagonals bisect]

Now as O and B are mid points of AC and AF . By mid point theorem we can say that CF || BORightarrow FG || BD

AS BD || FG and AC is the transversal

angle AOB = angle ACF               .................(1)          [Corresponding angle]

Similarly EG || AC and GF is the transversal

angle ACF = angle EGF               .................(2)          [Corresponding angle]

From (1) and (2) We get

angle EGF = angle AOB =90^0, Hence Proved

Sample Questions
(More Questions for each concept available in Login)
Question : 1

The line segment joining the mid-points of any two sides of a triangle is ____________ .

Right Option : D
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Explanation
Question : 2

D and E are the mid-points of sides AB and AC of DeltaABC, and O is any point on side BC. O is joined to A. If P and Q are th mid-points id OB and OC respectively, then DEQP is a _________________.

Right Option : D
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Explanation
Question : 3

In a Delta ABC,D,E,F are the mid-points of sides BC,CA;and;AB respectively. If ar(Delta ABC)=16;cm^2, then  ar(trapezium FBCE) =

Right Option : C
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Explanation
 
 
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