In the previous section we saw a new method to find the Circumcentre. In this section, we will see Medians.
We know that, in a triangle, a
median is a line drawn from a vertex, to the midpoint of the opposite
side. We have seen it before here and here. Let us now draw the three medians
of a triangle.
• Fig.18.32(a) shows a triangle ABC. A median is drawn from vertex C. That is., a line is drawn from the vertex C to the midpoint D of the opposite side BC.
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| Fig.18.32 |
• In fig.b, another median is
drawn from the vertex A.
• And finally, in fig.c, the third median is
drawn from vertex B.
■ Note that, the three medians pass through the
same point. We will call this point as ‘G’.
We will now do some calculations with these medians.
1. Consider any two medians. Let
us take those medians from A and B. This is shown in fig.18.33(a) below:
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| Fig.18.33 |
2. The ends of the medians are E
and F. They are joined by an yellow line.
3. E and F are the midpoints of
sides BC and AC. So, by theorem 18.5, EF will be half of AB. That is., EF = AB⁄2
4. Now, there is a smaller
triangle ABG inside. Mark the midpoints of it’s sides AG and BG.
Let those midpoints be P and Q. This is shown in fig.b
5. Join PQ. Since P and Q are
midpoints, by theorem 18.5, PQ will be half of AB. That is., PQ = AB⁄2
6. From (3) and (5), we get PQ = EF
7. We know that PQ is parallel to
AB, and EF is also parallel to AB. So PQ and EF are parallel to
each other.
8. Thus, in the quadrilateral PQEF
shown in fig.18.33(c) below, the opposite sides PQ and EF are equal and parallel.
So PQEF is a parallelogram.
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| Fig.18.33 |
9. In any parallelogram, the
diagonals will bisect each other. So in our parallelogram PQEF, the
diagonals PE and FQ will bisect each other. So we can write: • PG = GE • QG = GF
10. Now let us move back to our
original medians AE and BF shown in fig.a. They are now divided at
various points P, Q and G. This is shown in fig.d.
11. Consider the median AE. It is
divided into three parts AP, PG and GE.
(i) From (4), P is the midpoint of
AG. So AP = PG
(ii) From (11.i) and (9), we get: AP = PG =
GE. That means, the median AE is divided into three equal parts by
points P and G
(iii) On one side of G, there are
two equal parts AP and PG, and on the other side, there is one equal
part GE.
(iv) So G divides the median AE in
the ratio 2:1
The same result of 2:1 ratio
can be obtained for the other median BF also. The steps are similar
to those from to. But we will write them again:
12. Consider the median BF. It is
divided into three parts BQ, QG and GF.
(i) From (4), Q is the midpoint of
BG. So BQ = QG
(ii) From (12.i) and (9), we get: BQ = QG =
GF. That means, the median BF is divided into three equal parts by
points Q and G
(iii) On one side of G, there are
two equal parts BQ and QG, and on the other side, there is one equal
part GF.
(iv) So G divides the median BF in
the ratio 2:1
13. So we find that, G is an
important point. It divides the two medians in the ratio 2:1
■ Now, what is this G? Is it
very difficult to find the position of G?
Ans: • From fig.18.32(c), G is the point
of intersection of the medians of a triangle.
• It is not at all difficult to
find the position of G. All we need to do is, to draw any two
medians. Their point of intersection is the point ‘G’
14. In fig., we considered a
convenient pair of medians. AE and BF. We could take any of the other
two possible pairs: • AE and CD • BF and CD
15. In any case that we take, we
will get the same result:
■ The point of intersection of
the two medians will divide them both in the ratio 2:1
16. For example, if we take AE and
CD, we can write this:
(i) G divides the median AE in the
ratio 2:1
(ii) G divides the median CD in the
ratio 2:1
From 11, 12 and 16 , we can say that G divides all the three medians in the same ratio 2:1.
We will write it in the form of a theorem:
Theorem 18.6
The point of intersection G divides all the three medians in the ratio 2:1 measured from the vertex
The point of intersection G divides all the three medians in the ratio 2:1 measured from the vertex
Until now, we have been
discussing the 'action of parallel lines' in the interior portion of
triangles, and also on the sides of the triangles. Now we will have a short discussion on some thing which
happens out side a triangle. Fig.18.34(a) shows a triangle ABC. In
the fig.b, the sides AC and BC are extended upwards beyond the vertex
C.
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| Fig.18.34 |
In the fig.18.34(c) below, a line is
drawn parallel to the base AB, cutting through the extended portions.
Let it intersect the extensions at P and Q.
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| Fig.18.34 |
A red line parallel to AB
is drawn through C. This is shown in fig.d. We do not need the
portions beyond P and Q. So they are trimmed.
1. Thus, in fig.d, we have three
parallel lines: • AB, • 'the red line' and • PQ,
2. Those parallel lines cut through the lines AQ
and BP. The distances cut are in the same ratio (Theorem 18.1). So we
have: AC:QC = BC:PC ⇒ AC⁄QC = BC⁄PC .
3. Consider a different ratio: PB⁄PC .
PB⁄PC = (PC + BC)⁄PC (∵ PB = PC + BC)
⇒ PB⁄PC = PC⁄PC + BC⁄PC ⇒ PB⁄PC = 1 + BC⁄PC .
4. Similarly, consider: AQ⁄CQ .
AQ⁄CQ = (CQ + AC)⁄CQ (∵ AQ = CQ + AC)
⇒ AQ⁄CQ = CQ⁄CQ + AC⁄CQ ⇒ AQ⁄CQ = 1 + AC⁄CQ .
5. Let us rewrite the results in (3) and (4):
From (3) we have: PB⁄PC = 1 + BC⁄PC .
From (4) we have: AQ⁄CQ = 1 + AC⁄CQ .
6. The last terms in the above two equations are the same. This we know from (2)
7. So (3) and (4) are equal. That is., PB⁄PC = AQ⁄CQ .
■ We can write a summary of the
above discussion as follows:
• Extend the sides of a triangle
• Consider one side. Take the
ratio: (Extension)⁄(Total length)
• Consider the other side. Take
the ratio: (Extension)⁄(Total length)
• Both the ratios will be equal
Now we will see a solved example
Solved example 18.10
Solved example 18.10
In the fig.13.35 below, ABC is a right angled
triangle. D is the midpoint of the
hypotenuse AC. DE is drawn perpendicular to AB.
(i) Prove that DE is half of BC (ii) Prove that, in the larger ⊿ABC, the distances from D to all the vertices are equal (iii) Prove that D is the
circumcentre of ⊿ABC
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| Fig.18.35 |
Solution:
Part (i): 1. DE is parallel to
BC. Because, both DE and BC are perpendicular to AB. Also, D is the midpoint of AC.
2. Theorem 18.4 states: In any triangle, the line drawn parallel to one side, passing through the midpoint of another side, meets the third side also at it's midpoint.
■ So E is the midpoint of AB
3. Now we apply theorem 18.5: The length of a line joining the midpoints of two sides of a triangle is half the length of the third side.
■ So DE is half of BC
2. Theorem 18.4 states: In any triangle, the line drawn parallel to one side, passing through the midpoint of another side, meets the third side also at it's midpoint.
■ So E is the midpoint of AB
3. Now we apply theorem 18.5: The length of a line joining the midpoints of two sides of a triangle is half the length of the third side.
■ So DE is half of BC
Part (ii): The distances to
the three vertices are DA, DB and DC. We have to prove that, these
three distances are equal.
1. We already know that DA = DC.
Because D is the midpoint of AC
Now we will prove that DA =
DB. The proof is as follows:
2. In fig.18.35(b) above, consider the triangles: ⊿AED
and ⊿BED
• ED = ED (The common side)
• AE = BE (since E is the
midpoint of AB)
• ∠AED and ∠BED are colinear
angles, and one of them (∠AED) ie 90o. So ∠BED = 180 – 90 = 90o
• Thus, ∠AED = ∠BED = 90o
3. Thus we have two sides and
included angle same in both the triangles. It is a case of SAS congruence. The two triangles ⊿AED and ⊿BED are equal.
• Let us write the
correspondence:
• ED is the common side. The 90o is at E. So E↔E
and D↔D
• The remaining vertices are A
and B. So A↔B
• We can write: The correspondence of vertices is: E↔E, D↔D and A↔B
• From the above, we can pick
each corner and write the correspondence of sides: ED↔ED, EA↔EB and DA↔DB
• From the last one DA↔DB, we can
write DA= DB
4. From [Part (ii) 1], we have: DA = DC
• From [Part (ii) 3], we have DA = DB
• So we can write DA = DB = DC
5. Thus, the distances from D to all
the vertices of ⊿ABC are equal
Part (iii):
• If D is the circumcentre, there exists a circle, with D as it's centre, and with the vertices A, B and C lying on it.
• If D is the circumcentre, there exists a circle, with D as it's centre, and with the vertices A, B and C lying on it.
• Indeed there exists such a
circle because in part (ii), we proved that A, B and C are at
equidistance from D. SO D is the circumcentre of
ABC
We can prove this in another
way also:
1. Draw the perpendicular
bisector of BC as shown in fig.c below
2. Let it bisect BC at F. Since,
it is a bisector, BF is half of BC
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| Fig.18.35 |
3. Now, using theorem 18.5, DE is also
half of BC
4. That means., D and F, are at
the same perpendicular distance from the side AB. Also, the perpendicular
bisector through F will be parallel to AB
5. So, if we extend the
perpendicular bisector, it will pass through D. See fig.d
6. We already know that DE is
the perpendicular bisector of AB (since it is perpendicular to AB,
and E is the midpoint of AB)
7. So we have the perpendicular
bisectors of two sides intersecting at D. Thus D is the circumcentre
An even simpler method:
1. The perpendicular bisector of
AB, already passes through D
2. We need one more perpendicular
bisector.
3. Why not take the one of the
hypotenuse AC. It will pass through D itself.
4. So we get two perpendicular
bisectors, and both of them intersect at D. Thus D is the circumcentre
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