CBSE Class 9 > Class 9 Videos > L13 : Mid point theorem - Quadrilaterals, Maths, Class 9
L13 Mid point theorem - Quadrilaterals, Maths, Class 9 Video Lecture
FAQs on L13 : Mid point theorem - Quadrilaterals, Maths, Class 9
| 1. What is the mid point theorem in quadrilaterals? |  |
Ans. The midpoint theorem states that the line segment joining the midpoints of two sides of a quadrilateral is parallel to the other two sides and is half of their length.
| 2. How can the mid point theorem be applied to quadrilaterals? | |
Ans. To apply the midpoint theorem to quadrilaterals, we need to identify the midpoints of two sides and then draw a line segment joining these midpoints. This line segment will be parallel to the other two sides and will be half of their length.
| 3. Can the mid point theorem be used in any type of quadrilateral? | |
Ans. Yes, the mid point theorem can be used in any type of quadrilateral, including squares, rectangles, parallelograms, and trapezoids. As long as we can identify the midpoints of two sides, we can apply the theorem.
| 4. What is the significance of the mid point theorem in quadrilaterals? | |
Ans. The mid point theorem is significant in quadrilaterals as it helps us understand the relationship between the sides and diagonals of a quadrilateral. It also provides a method to prove that certain lines in a quadrilateral are parallel.
| 5. How can the mid point theorem be used to solve problems in quadrilaterals? | |
Ans. The mid point theorem can be used to solve problems in quadrilaterals by providing a way to prove that certain lines are parallel. This can help in finding the lengths of sides or diagonals, determining angles, or establishing congruence between different parts of the quadrilateral.
Text Transcript from Video
hello friends this video on
quadrilaterals part 13 is brought to you
by exam feel calm no more fear from exam
let's learn a very important concept
called my quadrature Allah quest you
guys to pause this video for some time
get a piece of pencil and pen and then
stop will do activity let's draw a
triangle and then let's mark the
midpoint of these sites mom this points
ABC it's tore the midpoint of a B and E
C these are the midpoints of a B and a C
and if we call this as e inner join this
point ENF and then you measure EF and bc
with scale you'll observe that EF will
always be equal to half of BC please
observe with this will always be true
okay
also you observe this angle and this
angle will always be and this angle and
this angle will also always be equal you
can just measure with a protractor you
will see that a EF and ABC is always
equal and EFC and ACB is always e so
three observations one s EF is equal to
half BC and these angles are equal if
this angles are equal that means they
are parallel right so that means EF will
always be parallel to BC so you can say
these are the observations if you draw a
line segment by the midpoint of these
two sides and these line is actually
parallel to and it says it is midpoint
they are midpoint and EF will be equal
to half of BC that is observed data here
to prove it I'll prove that but first
try find this on a piece of paper and
observe that you are getting EF is equal
to ha BC and you're also getting EF is
parallel to BC okay now since this is a
theorem let's try to prove it theorem L
is the line segment joining the
midpoints of the two sides of the
is parallel to the third side that is
this line segment running the midpoint
of the two sides of these triangles this
parallel to the third side that is this
is equal to this this is given so we
have to prove that yet is bothered too
easy so let me write this what is given
given this AE is equal to EB and AF is
equal to FC right to do what to prove f
is parallel to BC that is what we have
to be okay to prove this we can draw a
CD mrs. Palin to be a sinister Rossini
we draw this line extend this and then
this B D so draw CD pilot to be could be
you can see CD is parallel to B now we
see therefore e BC B is this a problem
you don't know you don't know because
that is what you do okay now if you see
if you take these two triangles this
triangle this triangle we can prove that
they are congruent so these two sides
are equal these two angles will be equal
okay and this is parallel to this so
these angles are equal
okay similarly this angle is also equal
so if you take these two triangles
normal hee-hee and see the F you see
that angle a is equal to angle C angle E
is equal to angle D if they have seen
particular was triangle also one side is
same that is AF and FC AF and so but
this if you prove that they're congruent
click see they're comprehending you say
that EF is equal to DF but this EF is
equal to B here also a is equal to C B
is equal to C D so now if a is equal to
CD and a is equal to b YZ v is equal to
c d and b is parallel to CD so by the
theorem if their lines are farther on
week but it's a parallelogram since it
is a parallelogram e f could be parallel
to BC let me say that in triangle a e f
and triangle C D F if you see let's take
one side let's take one like an e a d e
f equal to angle C D F or DC F if F is
equal to C here
and EF a is equal to angle DFA this
vertical opposite angle this is our
triangle and this is given therefore
triangle AE f is congruent to triangle
CDF since they are congruent
ae is equal to CD okay and EF is equal
to DF here and you say a is equal to CD
a is equal to CD instead of a I can say
what B plus a is equal to P given sheet
okay ECD I can say that he is equal to
CD now if P is equal to CD and EB is
parallel to CD he B is parallel to CD
construction it has been construction
there for ve DC's parallelogram EP Dec
zapallar am ii ii d has to be parallel
to BC origin of EDI can say that iam has
to be parallel to BC and that is what I
have to do okay also note that also mean
more one more thing this way it is not
required in this question but I'm gonna
show you this part so you know that EF
is equal to DF right you know so BC is
equal to what you need that is those of
you know website so paragraph right so
BC by 2 is 180 by T and what is DD by 2
that is here so what I am that is also
we can easily prove correct 20 BC is
equal to FD VC by 2 is equal to M D by 2
and FD y to the 1 here
okay let's see the converse is true
let's do an activity so let's draw
triangle okay and let's take a midpoint
of this side you see you take a midpoint
of the side a B in mark at this point E
now draw a parallel from here from you
draw lines parallel to BC okay so I'm
not drawing a midpoint here first off
obviously I'm just drawing a line that
is parallel to BC and which passes
through E and then that will meet
somewhere at some point so we have to we
will see that you can do this on your
paper you see that this line let's
suppose this is EF EF is parallel to BC
you can try so draw triangle and take
this point E and draw line valid to the
F draw a line parallel to BC pausing to
e you see that this line touches EC a
point F and then you know observe that
you'll observe that F is equal to FC
okay
so you you are drawing what you're
drawing point E this midpoint or maybe
you're drawn this Plus you ever drawn
yep travelling to BC originally you've
drawn some lines through here this
parallel to BC and this then you know
that this point F is nothing but
bisecting side easy that is all the way
let's see if we can prove it
this is a converse Hyrum see the
statement the line drawn through the
midpoint of one side and parallel to 1
one side pilot to another site by Sigma
third site this is the first time if the
second side is 30 it's also so lying
drawn from the midpoint of one side for
side parlor to the second side will
bisect dong third site that is what we
have to prove similarly we can prove in
the same fashion
draw this line okay and in this meet at
point B and do some podium so what we
are doing is we have drawn so we have
drawn see deeper into being or B or
whatever you want me to say okay and it
has given that a is equal to EB so it is
given a is equal to B so what we have to
prove and also EF is parallel to BC so
we have to prove that EF is equal to FC
okay so now if you take this triangle AE
air and see the degrees of triangles let
us take this triangle here and CBF let's
see what is common so AEMC be there in
addition a CD yes you see he actually
will be equal to C well because he is
going to be he and you have drawn this
in such a way that this is parallel
whites we have drawn a paddle paddle
line here and we have done a
parallelogram actually hey Zoe
ae is equal to B is equal to C so a is
producing okay then let's see 9s a EF a
here this
if these are parallel now the sign will
be equal to the same so a EF is equal to
FP c AE m will be equal to FD c correct
so any EF t will be equal to C
everything
what do you love is thing so now I'm you
see angle angle side with a yes they are
completely okay so this first mention
that triangle e EF is congruent to
triangle C D F my angle I can sign okay
now since they are congruent what can be
achieved from this say they are
confident we can say that can we say F
is equal to FC yes and that is that's
all you had to prove right that means F
is the midpoint to piece it easy okay
thank you visit our website exam viacom
to watch more and more qualities we
should be news you can also attempt free
online tests that are there in our
website you can also get access to tons
of free study materials and you can also
find free tutors and mentors in this
website thanks a lot for watching
quadrilaterals part 13 is brought to you
by exam feel calm no more fear from exam
let's learn a very important concept
called my quadrature Allah quest you
guys to pause this video for some time
get a piece of pencil and pen and then
stop will do activity let's draw a
triangle and then let's mark the
midpoint of these sites mom this points
ABC it's tore the midpoint of a B and E
C these are the midpoints of a B and a C
and if we call this as e inner join this
point ENF and then you measure EF and bc
with scale you'll observe that EF will
always be equal to half of BC please
observe with this will always be true
okay
also you observe this angle and this
angle will always be and this angle and
this angle will also always be equal you
can just measure with a protractor you
will see that a EF and ABC is always
equal and EFC and ACB is always e so
three observations one s EF is equal to
half BC and these angles are equal if
this angles are equal that means they
are parallel right so that means EF will
always be parallel to BC so you can say
these are the observations if you draw a
line segment by the midpoint of these
two sides and these line is actually
parallel to and it says it is midpoint
they are midpoint and EF will be equal
to half of BC that is observed data here
to prove it I'll prove that but first
try find this on a piece of paper and
observe that you are getting EF is equal
to ha BC and you're also getting EF is
parallel to BC okay now since this is a
theorem let's try to prove it theorem L
is the line segment joining the
midpoints of the two sides of the
is parallel to the third side that is
this line segment running the midpoint
of the two sides of these triangles this
parallel to the third side that is this
is equal to this this is given so we
have to prove that yet is bothered too
easy so let me write this what is given
given this AE is equal to EB and AF is
equal to FC right to do what to prove f
is parallel to BC that is what we have
to be okay to prove this we can draw a
CD mrs. Palin to be a sinister Rossini
we draw this line extend this and then
this B D so draw CD pilot to be could be
you can see CD is parallel to B now we
see therefore e BC B is this a problem
you don't know you don't know because
that is what you do okay now if you see
if you take these two triangles this
triangle this triangle we can prove that
they are congruent so these two sides
are equal these two angles will be equal
okay and this is parallel to this so
these angles are equal
okay similarly this angle is also equal
so if you take these two triangles
normal hee-hee and see the F you see
that angle a is equal to angle C angle E
is equal to angle D if they have seen
particular was triangle also one side is
same that is AF and FC AF and so but
this if you prove that they're congruent
click see they're comprehending you say
that EF is equal to DF but this EF is
equal to B here also a is equal to C B
is equal to C D so now if a is equal to
CD and a is equal to b YZ v is equal to
c d and b is parallel to CD so by the
theorem if their lines are farther on
week but it's a parallelogram since it
is a parallelogram e f could be parallel
to BC let me say that in triangle a e f
and triangle C D F if you see let's take
one side let's take one like an e a d e
f equal to angle C D F or DC F if F is
equal to C here
and EF a is equal to angle DFA this
vertical opposite angle this is our
triangle and this is given therefore
triangle AE f is congruent to triangle
CDF since they are congruent
ae is equal to CD okay and EF is equal
to DF here and you say a is equal to CD
a is equal to CD instead of a I can say
what B plus a is equal to P given sheet
okay ECD I can say that he is equal to
CD now if P is equal to CD and EB is
parallel to CD he B is parallel to CD
construction it has been construction
there for ve DC's parallelogram EP Dec
zapallar am ii ii d has to be parallel
to BC origin of EDI can say that iam has
to be parallel to BC and that is what I
have to do okay also note that also mean
more one more thing this way it is not
required in this question but I'm gonna
show you this part so you know that EF
is equal to DF right you know so BC is
equal to what you need that is those of
you know website so paragraph right so
BC by 2 is 180 by T and what is DD by 2
that is here so what I am that is also
we can easily prove correct 20 BC is
equal to FD VC by 2 is equal to M D by 2
and FD y to the 1 here
okay let's see the converse is true
let's do an activity so let's draw
triangle okay and let's take a midpoint
of this side you see you take a midpoint
of the side a B in mark at this point E
now draw a parallel from here from you
draw lines parallel to BC okay so I'm
not drawing a midpoint here first off
obviously I'm just drawing a line that
is parallel to BC and which passes
through E and then that will meet
somewhere at some point so we have to we
will see that you can do this on your
paper you see that this line let's
suppose this is EF EF is parallel to BC
you can try so draw triangle and take
this point E and draw line valid to the
F draw a line parallel to BC pausing to
e you see that this line touches EC a
point F and then you know observe that
you'll observe that F is equal to FC
okay
so you you are drawing what you're
drawing point E this midpoint or maybe
you're drawn this Plus you ever drawn
yep travelling to BC originally you've
drawn some lines through here this
parallel to BC and this then you know
that this point F is nothing but
bisecting side easy that is all the way
let's see if we can prove it
this is a converse Hyrum see the
statement the line drawn through the
midpoint of one side and parallel to 1
one side pilot to another site by Sigma
third site this is the first time if the
second side is 30 it's also so lying
drawn from the midpoint of one side for
side parlor to the second side will
bisect dong third site that is what we
have to prove similarly we can prove in
the same fashion
draw this line okay and in this meet at
point B and do some podium so what we
are doing is we have drawn so we have
drawn see deeper into being or B or
whatever you want me to say okay and it
has given that a is equal to EB so it is
given a is equal to B so what we have to
prove and also EF is parallel to BC so
we have to prove that EF is equal to FC
okay so now if you take this triangle AE
air and see the degrees of triangles let
us take this triangle here and CBF let's
see what is common so AEMC be there in
addition a CD yes you see he actually
will be equal to C well because he is
going to be he and you have drawn this
in such a way that this is parallel
whites we have drawn a paddle paddle
line here and we have done a
parallelogram actually hey Zoe
ae is equal to B is equal to C so a is
producing okay then let's see 9s a EF a
here this
if these are parallel now the sign will
be equal to the same so a EF is equal to
FP c AE m will be equal to FD c correct
so any EF t will be equal to C
everything
what do you love is thing so now I'm you
see angle angle side with a yes they are
completely okay so this first mention
that triangle e EF is congruent to
triangle C D F my angle I can sign okay
now since they are congruent what can be
achieved from this say they are
confident we can say that can we say F
is equal to FC yes and that is that's
all you had to prove right that means F
is the midpoint to piece it easy okay
thank you visit our website exam viacom
to watch more and more qualities we
should be news you can also attempt free
online tests that are there in our
website you can also get access to tons
of free study materials and you can also
find free tutors and mentors in this
website thanks a lot for watching
About this Video
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Apr 18, 2026 Last updated
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