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Mid Point Theorem

Quadrilaterals II - Concepts

Properties of Rectangle

Diagonals of a Rectangle Are of Equal Length

Properties of Rhombus

The Diagonal of a Rhombus are Perpendicular to Each Other

Properties of Square

Diagonals of Square Are Equal and Perpendicular to Each Other.

Mid Point Theorem

Converse Mid Point Theorem

Class - 9th IMO Subjects

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 || AO $Rightarrow$ 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 || BO $Rightarrow$ 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

ABCD is a parallelogram in which P,Q,R,S are mid-points of the sides AB, BC, CD and DA respectively. AC is a diagonal. Then which of the following is true?
A. $SR=frac{1}{2}AC$	B. $PQ=frac{1}{2}AC$	C. SR \|\| AC, PQ \|\| AC	D. All of these
Right Option : D
View Explanation

Question : 2

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) =
A. $4 ; cm^2$	B. $8 ; cm^2$	C. $12;cm^2$	D. $10;cm^2$
Right Option : C
View Explanation

Question : 3

In $Delta$ ABC, if D, F and E are the mid-points of the sides AB, AC and BC respectively, then which of the following relationships is true?
A. $angle$ BDF = $angle$ DBE	B. $angle$ DFE = $angle$ EFC	C. $angle$ DBE = $angle$ EFD	D. $angle$ BEF = $angle$ DFE
Right Option : C
View Explanation