You have seen above that if the three angles of one triangle are respectively equal to the
three angles of another triangle, then their corresponding sides are proportional
(i.e., in the same ratio). What about the converse of this statement? Is the converse true?
In other words, if the sides of a triangle are respectively proportional to the sides of another
triangle,
is it true that their corresponding angles are equal? Let us examine it through an activity :
Activity 5 : Draw two triangles ABC and DEF such that
AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm (see Fig.).
Given: ${AB}/{DE}={BC}/{EF}={CA}/{FD}=2/3$
Now measure ∠ A, ∠ B, ∠ C, ∠ D, ∠ E and ∠ F. You will observe that
∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F, i.e., the corresponding angles of the two triangles are equal.
You can repeat this activity by drawing several such triangles (having their sides in the same ratio).
Everytime you shall see that their corresponding angles are equal. It is due to the following
criterion of similarity of two triangles:
Theorem 6.4 : If in two triangles, sides of one triangle are proportional to
(i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles
are equal and hence the two triangles are similiar.
This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
${AB}/{DE}={BC}/{EF}={CA}/{FD}(<1)$ (see Fig.):
Cut DP = AB and DQ = AC and join PQ.
It can be seen that ${DP}/{PE}={DQ}/{QF}$ and PQ || EF
So, ∠ P = ∠ E and ∠ Q = ∠ F.
Therefore ${DP}/{DE}={DQ}/{DF}={PQ}/{EF}$
So, ${DP}/{DE}={DQ}/{DF}={BC}/{EF}$
So, BC = PQ
Thus, ∆ ABC ≅ ∆ DPQ
So ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠F
Remark : You may recall that either of the two conditions namely,
(i) corresponding angles are equal and
(ii) corresponding sides are in the same ratio is not sufficient for two polygons to be similar.
However, on the basis of Theorems 6.3 and 6.4, you can now say that in case of similarity of the
two triangles, it is not necessary to check both the conditions as one condition implies the other.
Let us now recall the various criteria for congruency of two triangles learnt in Class IX.
You may observe that SSS similarity criterion can be compared with the SSS congruency criterion.
This suggests us to look for a similarity criterion comparable to SAS congruency criterion of triangles.
For this, let us perform an activity.
Activity 6 : Draw two triangles ABC and DEF such that AB = 2 cm, ∠ A = 50°, AC = 4 cm,
DE = 3 cm, ∠ D = 50° and DF = 6 cm (see Fig).
Here, you may observe that ${AB}/{DE}={AC}/{DF} $ each equal to $2/3$ and ∠ A (included
between the sides AB and AC) = ∠D (included between the sides DE and DF).
That is, one angle of a triangle is equal to one angle of another triangle and sides including these
angles are in the same ratio (i.e., proportion). Now let us measure ∠ B, ∠ C, ∠ E and ∠ F.
You will find that ∠B = ∠E and ∠C = ∠F. That is, ∠A = ∠D, ∠B = ∠E and
∠ C = ∠ F. So, by AAA similarity criterion, ∆ ABC ~ ∆ DEF. You may repeat this activity by drawing
several pairs of such triangles with one angle of a triangle equal to one angle of another triangle
and the sides including these angles are proportional. Everytime, you will find that the triangles
are similar. It is due to the following criterion of similarity of triangles:
Theorem 6.5 : If one angle of a triangle is equal to one angle of the other triangle
and the sides including these angles are proportional, then the two triangles are similar.
Proof: This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.
As before, this theorem can be proved by taking two triangles ABC and DEF such that
${AB}/{AC}={DE}/{DF}(<1)$ and ∠ A = ∠ D .(see Fig.).
Cut DP = AB, DQ = AC and join PQ.
Now, PQ || EF and ∆ ABC ≅ ∆ DPQ
So, ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q
Therefore, ∆ ABC ~ ∆ DEF
We now take some examples to illustrate the use of these criteria.
Example 4 : In Fig. 6.29, if PQ || RS, prove that ∆ POQ ~ ∆ SOR.
Solution : PQ || RS (Given)
So, ∠ P = ∠S (Alternate angles)
and ∠ Q = ∠ R
Also, ∠ POQ = ∠ SOR (Vertically opposite angles)
Therefore, ∆ POQ ~ ∆ SOR (AAA similarity criterion)
Example 5 : Observe in Fig. and then find ∠ P.
Solution : In ∆ABC and ∆ PQR,
${AB}/{RQ}=3.8/7.6=1/2$;${BC}/{QP}=6/12=1/2$ and ${CA}/{PR}={3√3}/{6√3}=1/2$
That is, ${AB}/{RQ}={BC}/{QP}={CA}/{PR}$
So, ∆ ABC ~ ∆ RQP (SSS similarity)
Therefore, ∠ C = ∠ P (Corresponding angles of similar triangles)
But, ∠ C = 180° – ∠ A – ∠ B (Angle sum property)
= 180° – 80° – 60° = 40°
So, ∠ P = 40°
Example 6 : In Fig. 6.31, OA . OB = OC . OD. Show that ∠ A = ∠ C and ∠ B = ∠ D.
Solution : OA . OB = OC . OD (Given)
So, ${OA}/{OC}={OD}/{OB} .............(1)$
Also, we have ∠ AOD = ∠ COB (Vertically opposite angles).....(2)
Therefore, from (1) and (2), ∆ AOD ~ ∆COB (SAS similarity criterion)
So, ∠ A = ∠ C and ∠ D = ∠ B (Corresponding angles of similar triangles)
Example 7 : A girl of height 90 cm is walking away from the base of a
lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her
shadow after 4 seconds.
Solution : Let AB denote the lamp-post and CD the girl after walking for 4 seconds away
from the lamp-post (see Fig. 6.32).
From the figure, you can see that DE is the shadow of the girl. Let DE be $x$ metres.
Now, BD = 1.2 m × 4 = 4.8 m.
Note that in ∆ ABE and ∆ CDE,
∠ B = ∠ D (Each is of 90° because lamp-post as well as the girl are standing vertical to the ground)
and ∠ E = ∠ E (Same angle)
So, ∆ABE ~ ∆ CDE (AA similarity criterion)
Therefore, ${BE}/{DE}={AB}/{CD}$
i.e ${4.8+x}/{x}=3.6/0.9$ $(90 cm)=90/100$ m=0.9 m
i.e $4.8 + x =4x$
i.e $3x=4.8$
i.e $x=1.6$
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
Example 8 : In Fig. 6.33, CM and RN are respectively the medians of ∆ ABC and
∆ PQR. If ∆ABC ~ ∆ PQR, prove that :
(i) ∆ AMC ~ ∆ PNR
(ii) ${CM}/{RN}={AB}/{PQ}$
(iii) ∆ CMB ~ ∆ RNQ
Solution : (i) ∆ABC ~ ∆ PQR ..........(Given)
So, ${AB}/{PQ}={BC}/{QR}={CA}/{RP} .........(1)$
and ∠A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R .........(2)
But AB = 2 AM and PQ = 2 PN (As CM and RN are medians)
So, from (1), ${2AM}/{2PN}={CA}/{RP}$
i.e ${AM}/{PN}={CA}/{RP} .....................(3)$
Also, ∠ MAC = ∠ NPR ...........[From (2)] .... (4)
So, from (3) and (4),
∆ AMC ~ ∆ PNR (SAS similarity)......hence proved (5)
(ii) From (5), ${CM}/{RN}={CA}/RP ...............(6)$
But, ${CA}/{RP}={AB}/{PQ} .......... [From (1)]... (7)$
Therefore, ${CM}/{RN}={AB}/{PQ}$ .... [From (6) and (7)] (8)
(iii) ${AB}/{PQ}={BC}/{QR} ..........[From (1)] (8)$
Again, ${CM}/{RN}={BC}/{QR} ......[From (8)] (9)$
Therefore, ${CM}/{RN}={AB}/{PQ}-{2BM}/{2QN}$
Also, ${CM}/{RN}={BM}/{QN} ................(10)$
i.e ${CM}/{RN}={BC}/{QR}={BM}/{QN}............[From (9) and (10)]$
Therefore, ∆ CMB ~ ∆ RNQ (SSS similarity)
[Note : You can also prove part (iii) by following the same method as used for proving part (i).]
