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Chapter-11 MID POINT THEOREM
MATH CLASS-9
Module Objectives
•
•
•
•
State the mid-point theorem.
Prove mid-point theorem logically.
State the converse of mid-point theorem.
Solve problems and riders based on the mid-point theorem.
Analysis of Mid-Point Theorem
Construct the triangles as given below. D and E are the mid-points of the sides of AB
and AC of the ∆ABC.
(i)
(ii)
(iii)
Measure the lengths of DE and BC and the magnitudes of ABC and ADE in each
case and complete the table given in the following slide.
Analysis of Mid-Point Theorem
FIG NO.
DE
BC

ADE

ABC
i)
ii)
iii)
With the help of the above table answer the following questions:
1. What is the relation between the lengths of DE and BC?
2. What is the relation between  ADE and  ABC ?
3. What type of angles are  ADE and  ABC ?
4. Hence what type of lines are DE and BC ?
From the measurements you will observe that,
DE is half of BC. i.e. DE = 1/ 2 (BC).The corresponding angles ADE and ABC are
equal.
Hence , DE || BC.
Proof of Mid-Point Theorem
•
Theorem : The line joining the mid-points of any two sides of a triangle is parallel to and half
of the third side.
Data : In ∆ABC , D is the midpoint of AB and E is the
midpoint of AC
To Prove: (i) DE ||BC
(ii) DE = ½ BC
Analysis:
On what basis can we prove lines are parallel ?
(1) Corresponding or alternate angles are equal.
(2) The lines form the opposite sides of a
parallelogram.
Proof of Mid-Point Theorem
•
•
•

We not only have to prove DE || BC but also DE = ½ BC .How can we do this ?
This is possible by constructing a parallelogram which has BC and DF as its sides.
How can we show that DBCF is a parallelogram?
This is possible by proving DB = CF and DB ||CF
By constructing CF || DB.
We cannot prove that DB = CF directly.But we know DB = DA.
How can we prove that DB = CF ?
So we will prove DA = CF
Hence prove DB = CF by proving ∆ ADE ≡ ∆ EFC.
How can we prove that ∆ ADE ≡ ∆ EFC ?
By applying A.S.A postulate.
Proof of Mid-Point Theorem
Hence ,in order to prove DE || BC and DE = ½ BC logically, we need to
a) Construct CF || AB
b) Prove ∆ ADE ≡ ∆ EFC.
c) Prove DBCF is a parallelogram.
Theorem : The line segment joining the midpoints of any two sides of a triangle is
parallel to the third side and is equal to half of it.
Data : In ∆ ABC , D is the midpoint of AB and E is the midpoint of AC.
Proof of Mid-Point Theorem
To prove : DE || BC and DE = ½ BC.
Construction : From C, draw a line parallel to BA. Produce DE to meet this line at F.
STATEMENT
In ∆ADE and ∆ EFC, AE = EC
 AED

=
DAE =
 CEF
ECF
REASON
given
Vertically opposite angles.
Alternate angles. DA || CF
Hence , ∆ ADE ≡ ∆ EFC.
By A.S.A postulate
Hence , AD = CF and DE = EF
Corresponding parts of congruent ∆s.
But AD = BD
given., D is the midpoint of AB
Therefore, CF = BD
From the above two statements
Thus, DBCF is a parallelogram.
Opposite sides of CF and BD are equal
and parallel.
So, DE || BC
Opposite sides of a parallelogram are
parallel.
Proof of Mid-Point Theorem
STATEMENT
REASON
Now DE = EF
∆ADE ≡ ∆ CEF
DE = ½ BC
In parallelogram DBCF, DF = BC
Therefore , DE || BC and DE = ½ BC.
Hence , it is proved that, the line segment joining the mid points of any two
sides of a triangle is parallel to the third side and is equal to half of it.
Note: Prove this theorem by changing the construction as : Produce DE to F
such that EF = DE.
Converse of the theorem :
The straight line drawn through the midpoint of one side of a triangle and
parallel to another , bisects the third side.
Activity: a) Prove the above statement logically.
b) Verify it by practical method.
Examples
Example 1 : In the given figure , AB = 8.4 cm ,PR = 5.0 and PQ = 4.8cm .Find the length
of BC,CA and QR. If  APR =45⁰, find  ABC.
Given : AB = 8.4cm
PR = 5.0 cm
PQ = 4.8 cm
Solution :
(1) BC = 2 * PR (Theorem 3)
(2) CA = 2 * PQ (Theorem 3)
BC = 2 * 5
CA = 2 * 4.8
BC = 10 cm
CA = 9.6 cm
(3) QR = ½ AB
(Theorem 3)
(4)  B = APR = 45⁰
QR = ½ * 8.4
QR = 4.2 cm
Therefore, PR || BC , corresponding angles are equal.
Examples
Example 2 : Prove that the figure obtained by joining the mid-points of the adjacent
Sides of a quadrilateral ABCD.
To Prove: PQRS is a parallelogram.
Construction : Join BD.
Proof:
In ∆ABD, PS || BD and PS = ½ BD {Since, line joining the midpoints of two sides of a
∆ is parallel to and half the third side.}
In ∆BCD, QR || BD and QR = ½ BD
Therefore, PS || QR, PS = QR.
Hence PQRS is a parallelogram.
Do it Yourself
1) In ∆ ABC , D is mid-point of AB and E is mid point of BC. Calculate :
i) DE , if AC = 6.4 cm.
ii) DEB , if ACB = 63⁰
2.In ∆ PQR , A,B,C are the midpoints of sides PQ,QR and RP. If
the perimeter of ∆ PQR = 32 cm ,what is the perimeter of
∆ ABC ?
3) In the figure, X and Y are the midpoints of AB and AC respectively. Given that BC =
6cm.AB = 5.4 cm and AC = 5.0 cm. Calculate :
i) The perimeter of ∆ AXY.
ii) The perimeter of trapezium BCYX.
Do it Yourself
4) In ∆ ABC ,’Q’ and ‘M’ are the midpoints of AB and AC respectively. P and L are the
midpoints of AQ and AM respectively .If BC = 12 cm , find the length of PL.
5) Construct ∆ ABC, given BC = 7cm, AB = 8 cm and AC = 6cm . Mark D and E the
midpoints of AB and AC respectively. Construct parallelograms DBCF and ECBG.
6) Prove that the four triangles obtained by joining the midpoints of the sides of a
triangle are congruent.
7) Prove that the figure obtained by the joining the midpoints of the adjacent sides
of a rhombus is a rectangle.
8) In the given figure, PQRS is a parallelogram .
A and B are the midpoints of sides SP and SR
respectively .Prove that the area of ∆APR is
equal to area of ∆BPR.
Do it Yourself
9) Use the following figure to prove that : ( AB || PR || DC )
i) R is the mid-point of BC.
ii) PR = ½ ( AB + DC )
10) D,E and F are the mid-points of the sides AB,BC and CA of an isosceles ∆ ABC in
which AB = BC. Prove that ∆ DEF is also isosceles.
END OF CHAPTER

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