In the previous section we completed the discussion on Parallel lines and Triangle division. In this section, we will see Similarity of Triangles.
• Consider the ΔABC in fig.19.1(a). It has sides AB = 7 cm, AC = 4 cm and BC = 6 cm.
• We want a smaller ΔA’B’C’ in which the longest side is half the longest side of ΔABC
Is there any easier way to do this, with out actually measuring the angles? To find that, let us try a different method:
We will consider one particular line through P: The one, which is parallel to side BC. This is shown in fig.c
Some questions will arise at this stage of our discussion:
• Consider the ΔABC in fig.19.1(a). It has sides AB = 7 cm, AC = 4 cm and BC = 6 cm.
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| Fig.19.1 |
• The longest side in ΔABC is AB
which is 7 cm in length. So in the new triangle, the longest side
will be 3.5 cm long. With this information, let us draw A’B’C’. It is shown in the above fig.19.1(b).
1. Draw a horizontal line A’B’
3.5 cm long. Measure ∠CAB (coloured in green) from the original
triangle.
2. Mark the same angle in the new fig. in such a way that ∠C’A’B’ = ∠CAB
3. Measure ∠ABC (coloured in
white) from the original triangle. Mark the same angle in the new
fig. in such a way that ∠A’B’C’ = ∠ABC
4. Thus we get a new smaller ΔA’B’C’
Is there any easier way to do this, with out actually measuring the angles? To find that, let us try a different method:
In fig.19.2(b) below, the midpoint
of the longest side AB is marked as P. So AP = BP = 3.5 cm
We can draw numerous lines
through P. Two such lines PQ1 and PQ2 are shown in the fig.b
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| Fig.19.2 |
■ What will happen if we use the
line PQ1?
• We will get a new triangle
APQ1 alright. But two angles will be different:
♦ ∠APQ1 will be less
than the original ∠ABC.
♦ ∠AQ1P will be greater than the
original ∠ACB.
• So we cannot use the line PQ1
■ What will happen if we use the
line PQ2?
• We will get a new triangle
APQ2 alright. But two angles will be different.
♦ ∠APQ2 will be greater
than the original ∠ABC.
♦ ∠AQ2P will be less than the original ∠ACB.
• So we cannot use the line PQ2
We will consider one particular line through P: The one, which is parallel to side BC. This is shown in fig.c
This line meets AC at Q. It
has two properties:
• It passes through the midpoint
P of side AB
• It is parallel to side BC
So, applying theorem 18.5
which we saw in the previous chapter,
• PQ is half of BC
• Q is the midpoint of side AC
• Side PQ is half of side
BC
• Side AQ is half of side AC
■ Most importantly, all the
angles are same. How do we know that angles are same?
• PQ and BC are two
parallel lines cut by a transversal AB. So, the white angles ∠APQ = ∠ABC (∵ they are corresponding angles)
• PQ and BC are two parallel
lines cut by a transversal AC. So, the blue angles ∠AQP = ∠ACB (∵ they are corresponding angles)
• The green ∠CAP remains
the same.
So we have successfully
reduced the original triangle ABC to half
Let us write a summary of what
we have done:
1. We have an original ΔABC
2. We marked the midpoint P of
the longest side AB
3. Then we drew PQ parallel to
side BC
■ Just by doing the above two
steps we acheived:
• A new triangle with same
angles but sides exactly half
Some questions will arise at this stage of our discussion:
■ In the above example, we
marked point P such that AP is 1⁄2 of AB.
Instead of 1⁄2, if we mark
point P in such a way that, AP is 1⁄3 of AB, will we get all new
sides 1⁄3 of the corresponding original sides?
OR, in general, if we mark
point P in such a way that AP is x⁄y of AB, will we get all new sides x⁄y of the corresponding original sides? (Note that x⁄y can be any fraction like 1⁄4, 1⁄7, 3⁄11, ....)
■ Another question that arises:
Is it necessary to always mark
point P on the longest side AB itself? Can’t we mark P on any of
the sides to get the required new triangle?
The discussion below will give answers to the above questions:
The discussion below will give answers to the above questions:
1. Consider the ΔABC in
fig.19.3(a) The lengths of the sides are also marked.
• The length of the side which
is opposite to vertex A is ‘a’ cm
• The length of the side which
is opposite to vertex B is ‘b’ cm
• The length of the side which
is opposite to vertex C is ‘c’ cm
2. Fig.b shows a ΔPQR
which is obtained by shrinking ΔABC
 |
| Fig.19.3 |
(i) Assume that, side PQ is obtained by
shrinking side AB by a factor x⁄y. That means PQ = AB × x⁄y.
But PQ is given as ‘r’ cm. And AB is given as ‘c’ cm.
So we get r = c × x⁄y ⇒ r⁄c = x⁄y ⇒ c⁄r = y⁄x
But PQ is given as ‘r’ cm. And AB is given as ‘c’ cm.
So we get r = c × x⁄y ⇒ r⁄c = x⁄y ⇒ c⁄r = y⁄x
(ii) Assume that, side QR is obtained by
shrinking side BC by a factor u⁄v . That means QR = AC × u⁄v.
But QR is given as ‘p’ cm. And BC is given as ‘a’ cm.
So we get p = a × u⁄v ⇒ p⁄a = u⁄v ⇒ a⁄p = v⁄u
But QR is given as ‘p’ cm. And BC is given as ‘a’ cm.
So we get p = a × u⁄v ⇒ p⁄a = u⁄v ⇒ a⁄p = v⁄u
(iii) Assume that side PR is obtained by
shrinking side AC by a factor m⁄n. That means PR = AC × m⁄n.
But PR is given as ‘q’ cm. And AC is given as ‘b’ cm.
So we get q = b × m⁄n ⇒q⁄b = m⁄n ⇒b⁄q = n⁄m
But PR is given as ‘q’ cm. And AC is given as ‘b’ cm.
So we get q = b × m⁄n ⇒q⁄b = m⁄n ⇒b⁄q = n⁄m
■ Note that though the sides are
shrunk, angles remain unchanged. This is a very important fact in our discussion.
Now we begin the actual
calculations:
3. We need to compare
corresponding sides of the two triangles. For that,
• on the side AB, mark a point Q' at a distance of r cm from A. (see fig.c). This gives [AQ' = r] & [BQ' = c-r]
• on the side AC, mark a point R' at a distance of q cm from A. This gives [AR' = q] & [CR' = b-q]
• join Q' and R'
• join Q' and R'
4. Now we have a small ΔAQ'R' inside ΔABC
■ If we can prove that Q'R' is parallel to BC, we will get some very interesting results. But how do we prove it? Let us try:
5. Compare the triangles ΔAQ'R' and ΔPQR
• In ΔPQR, we have
(i) a side PQ of length r,
(ii) a side PR of length q
(iii) included angle between them ∠RPQ (coloured in green)
• In AQ'R', we have
(i) a side AQ' of length r,
(ii) a side AR' of length q
(iii) included angle between them ∠R'AQ' (coloured in green)
■ It is a case of SAS congruence. The two triangles are equal. Let us write the correspondence:
■ If we can prove that Q'R' is parallel to BC, we will get some very interesting results. But how do we prove it? Let us try:
5. Compare the triangles ΔAQ'R' and ΔPQR
• In ΔPQR, we have
(i) a side PQ of length r,
(ii) a side PR of length q
(iii) included angle between them ∠RPQ (coloured in green)
• In AQ'R', we have
(i) a side AQ' of length r,
(ii) a side AR' of length q
(iii) included angle between them ∠R'AQ' (coloured in green)
■ It is a case of SAS congruence. The two triangles are equal. Let us write the correspondence:
• In the SAS congruence that we
derived, the vertices P and A are our pivot points. So P↔A
• In ΔPQR, the vertex Q is at a
distance of r from the vertex P. In ΔAQ’R’, the vertex Q’ is at the same distance of r from A. So we get Q↔Q’
• In ΔPQR, the vertex R is at a
distance of q from the vertex P. In ΔAQ’P’, the vertex R’ is at the same distance of q from A. So we get R↔R’
6. Since R↔R', we can mark ∠AR’Q’ with blue colour, and
since Q↔Q', we can mark ∠AQ'R
with white colour. This is shown in fig.19.4(a) below:
 |
| Fig.19.4 |
7. Consider the two white angles
in fig.19.4(a). They indicate that the lines Q’R’ and BC make the same angle with the line AB. So Q’R’ and BC are parallel.
Thus we proved (4). [Note that this can be proved
with the blue angles also. CB and R’Q’ makes the same angle with
AC. So CB and R’Q’ are parallel]
8. So we come to the interesting
part:
Two parallel lines Q’R’
and BC cuts two lines AB and AC. Applying theorem 18.1, the distances
cut are in the same ratio.
9. We can write: (b-q)⁄q = (c-r)⁄r ⇒ b⁄q - q⁄q = c⁄r - r⁄r ⇒ b⁄q - 1 = c⁄r - 1 . ⇒ b⁄q = c⁄r
10. Now let us revisit step (3)
• We marked distances r and q
from the vertex A. In the same way, we can mark distances p and q from the
vertex C. Then we will get:
• Two parallel lines P’Q’
and AB cut through two lines AC and BC. This is shown in fig.19.4(b)
• Applying theorem 18.1 we get: (b-q)⁄q = (a-p)⁄p ⇒ b⁄q - q⁄q = a⁄p - p⁄p ⇒ b⁄q - 1 = a⁄p - 1 ⇒ b⁄q = a⁄p
11. From (9) and (10) we get: b⁄q = c⁄r = a⁄p .
12. But from 2(i), we have: c⁄r = y⁄x .
from 2(ii) we have a⁄p = v⁄u .
from 2(iii) we have b⁄q = n⁄m .
13. Comparing the above three with (11), we get: b⁄q = c⁄r = a⁄p = y⁄x = v⁄u = n⁄m .
14. From this we get: y⁄x = v⁄u = n⁄m ⇒ x⁄y = u⁄v = m⁄n .
15. That means, all the scaling factors that we applied on the sides of the original ΔABC are equal.
So we can say: If the new triangle has all the angles same as the original triangle, all the three sides of the original triangle was scaled by the same factor.
16. Since the three factors are the same, instead of writing x⁄y , u⁄v etc., we can give a common name to the three factors. Let this common name be 'k'
17. Then we can write:
• New smallest side = k × original smallest side ⇒ New smallest side⁄Original smallest side = k
• New medium side = k × original medium side ⇒ New medium side⁄Original medium side = k
• New longest side = k × original longest side ⇒ New longest side⁄Original longest side = k
18. Since all the three ratios are equal to 'k', we can write:
New smallest side⁄Original smallest side = New medium side⁄Original medium side = New longest side⁄Original longest side = k
We can write it in the form of
a theorem. We will write it in steps.
Theorem 19.1:
k is called the Scale factor.
Theorem 19.1:
If we have a triangle in hand,
and
• If we apply a scale factor
which is less than 1, the new triangle will be smaller
• If we apply a scale factor
which is greater than 1, the new triangle will be larger 
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