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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

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,
A. ABCD is a rhombus		B. Diagonals of ABCD are equal
C. Diagonals of ABCD are equal and perpendicular		D. Diagonals of ABCD are perpendicular.
Right Option : C
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

Chapters

Polynomials II

Direct and Inverse Variation

Linear Equations in One Variable

Introduction to Trigonometry

Introduction to Euclids Geometry

Herons Formula Complete

Polynomial I

Surface Area and Volume II

Surface Area and Volume I

Coordinate Geometry

Constructions

Areas Of Parallelograms And Triangles II

Areas Of Parallelograms And Triangles I

Linear Equations in Two Variables

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Navodaya Entrance 6th & 9th

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Attention !!!!!

Mid Point Theorem

Mid Point Theorem

Students / Parents Reviews [20]

Eshan Arora

Yogansh Nyasi

Barkha Arora

Sharandeep Singh

Archit Segal

Om Umang

Prisha Gupta

Manya

Cheshta

Anika Saxena

Upmanyu Sharma

Aman Kumar Shrivastava

Yatharthi Sharma

Aman Kumar Shrivastava

Hiya Gupta

Manish Kumar

Shreya Shrivastava

Ayan Ghosh

Ayan Ghosh

Hridey Preet