Similar Triangles | Criteria for Similarity of Two Triangles
Two triangles are said to be similar if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportional).
Thus, two triangles ABC and
(i)
(ii)
In this section, we shall make use of the theorems discussed in earlier sections to derive some criteria for similar triangles which in turn will imply that either of the above two conditions can be used to define the similarity of two triangles.
CHARACTERISTIC PROPERTY 1 (AAA SIMILARITY)
Theorem: If in two triangles, the corresponding angles are equal, then the triangles are similar.
Given:
Two triangles ABC and DEF in which
To prove:
DABC ~ DDEF
Proof:
Case 1: When AB = DE
In triangles ABC and DEF, we have
AB = DE [Given]
\
Þ BC = EF and AC = DF [c.p.c.t.]
Thus,
Hence,
Case 2: When AB < DE
Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ.
In DABC and DDPQ, we have
AB = DP [By construction]
ÐA = ÐD [Given]
AC =DQ [By construction]
\ DABC
\ ÐABC = ÐDPQ [c.p.c.t.] …(i)
But ÐABC = ÐDEF [Given] …(ii)
\ ÐDPQ = ÐDEF [c.p.c.t.]
But ÐDPQ and ÐDEF are corresponding angles.
Þ PQ || EF
\
\
Similarly
\
Hence, DABC ~ DDEF.
Case 3: When AB > DE
Let P and Q be points on AB and AC respectively such that AP = DE and AQ = DF. Join PQ.
In DAPQ and DDEF, we have
AP = DE [By construction]
AQ = DF [By construction]
ÐA = ÐD [Given]
\ DAPQ
\ ÐAPQ = ÐDEF [c.p.c.t.] …(i)
But ÐDEF = ÐABC [Given] …(ii)
From (i) and (ii) we have
\ ÐAPQ = ÐABC
But ÐAPQ and ÐABC are corresponding angles
\ PQ || BC
\
\
Similarly,
Thus,
or
Hence, DABC ~ DDEF.
CHARACTERISTIC PROPERTY 2 (SSS SIMILARITY)
Theorem: If the corresponding sides of two triangles are proportional, then they are similar.
Given:
Two triangles ABC and DEF such that
To prove: DABC ~ DDEF
Construction: Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ
Proof:
Þ
[
In DDEF, we have
\ PQ || EF [By the converse of BPT]
\ ÐDPQ = ÐDEF and ÐDQP = ÐDFE [Corresponding angles]
\ DDPQ = ~ DDEF [By AA similarity] …(ii)
\
or
But
From equations (iii) and (iv), we have
Þ BC = PQ …(v)
In DABC and DDPQ, we have
AB = DP [By construction]
AC = DQ [By construction]
BC = PQ [By (v)]
\ DABC
Þ DABC ~ DDPQ [
DABC ~ DDPQ] … (vi)
From equation (ii) and (vi), we get
DABC ~ DDEF
CHARACTERISTIC PROPERTY 3 (SAS SIMILARITY)
Theorem 5: If one angle of a triangle is equal to one angle of the other and the sides including these angles are proportional than the two triangles are similar.
Given: Two triangle ABC and DEF such that ÐA = ÐD and
To prove:
DABC ~ DDEF
Construction: Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ.
Proof: In DABC and DDPQ, we have
AB = DP [By construction]
AC = DQ [By construction]
ÐA = ÐD [Given]
\ DABC @
DDPQ [By SAS congruency] …(i)
Now,
\
In DDEF, we have
\ PQ || EF [By the converse of BPT]
\ ÐDPQ = ÐDEF and ÐDQP = ÐDFE [Corresponding angles]
\ DDPQ ~ DDEF [By AA similarity] …(iii)
From equations (i) and (iii), we get
DABC ~
DDEF.
Question:
Examine each pair of triangles in figure and state which pair of triangles are similar. Also, state the similarity criterion used and write the similarity relation in symbolic form.
Solution:
(i)
In triangles ABC and PQR, we observe that
ÐA = ÐQ = 40°, ÐB = ÐP = 60° and ÐC = ÐR = 80°
Therefore, by AAA-criterion of similarity
DBAC ~ DPQR
(ii)
In triangle PQR and DEF, we observe that
Therefore, by SSS-criterion of similarity, we have
DPQR ~ DDEF
(iii)
In D‘s MNP and EFG, we observe that
Therefore, these two triangles are not similar as they do not fulfill SSS-criterion of similarity.
(iv)
In D‘s DEF and MNP, we have
ÐD = ÐM = 70°
ÐE = ÐN = 80° [
So, by AA-criterion of similarity DDEF ~ DMNP.
Question:
Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
Solution:
Given:
DABC and DPQR in which BC = a, CA = b, AB = c and QR = p, RP = q, PQ = r.
Also, DABC ~ D PQR.
To Prove:
Proof:
Since DABC and DPQR are similar, therefore their corresponding sides are proportional.
Let
Þ a = kp, b = kq and c = kr
\
=
From (i) and (ii), we get
[each equal to k]
Question:
A vertical stick 12m long casts a shadow 8m long on the ground. At the same time a tower casts the shadow 40m long on the ground. Determine the height of the tower.
Solution:
Let AB be the vertical stick and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF.
Let DE = x metres.
We have,
AB = 12m, AC = 8m, and DF = 40m.
In DABC and DDEF, we have
ÐA = ÐD = 90° and ÐC = ÐF [Angle of elevation of the sun]
Therefore, by AA-criterion of similarity
DABC ~ DDEF
Þ
Þ
Question:
In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ^
BC and EF ^ AC, prove that DABD ~ ECF.
Solution:
Given:
An isosceles DABC in which AB = AC, E is a point on CB produced, AD ^ BC and EF ^ AC.
To prove:
DABD ~ DECF
Proof:
In DABC, since AB = AC,
therefore ÐC = ÐB [Ðs opposite to equal sides are equal]
In DABD and DECF
ÐB = ÐC [Proved above]
and ÐEFC = ÐADB [each 90°]
\ DABD ~ ECF [AA similarity]
Question:
E is a point on side AD produced of a parallelogram ABCD and BE intersects CD at F. Prove that DABE ~ DCFB.
Solution:
Given:
E is a point on side AD produced of a parallelogram ABCD and BE intersects CD at F
To prove:
DABE ~ DCFB
Proof: In D‘s ABE and CFB, we have
ÐAEB = ÐCBF [Alternate angles]
ÐA = ÐC [Opposite angles of a parallelogram]
\ DABE ~ DCFB. [By AA similarity]
Question:
In figure, if
Solution:
Given:
To prove:
DPQS ~ DTQR
Proof: We have,
Þ
We also have,
Ð1 = Ð2 [Given]
Þ PR = PQ [Sides opposite to equal angles are equal] …(ii)
From (i) and (ii), we get
Þ
Thus, in triangles PQS and TQR, we have
\ DPQS ~ DTQR. [By SAS similarity]
Question:
In figure, if DABE @ DACD, prove that DADE ~ DABC.
Solution:
Given: DABE @ DACD
To prove:
DADE ~ DABC
Proof: Since DABE @ DACD
\ AB = AC [cpct]…. (i)
and, AD = AE …. (ii)
Also,
Þ
Thus, in triangles ADE and ABC, we have
and, ÐBAC = ÐDAE [Common]
Hence, by SAS criterion of similarity.
DADE ~ DABC
Question:
In figure, AD and CE are two altitudes of DABC. Prove that
(i) DAEF ~ DCDF
(ii) DABD ~ DCBE
(iii) DAEF ~ DADB
(iv) DFDC ~ DBEC
Solution:
Given: AD and CE are two altitudes of DABC
To prove:
(i) DAEF ~ DCDF
(ii) DABD ~ DCBE
(iii) DAEF ~ DADB
(iv) DFDC ~ DBEC
Proof:
(i)
In triangles AEF and CDF, we have
ÐAEF = ÐCDF = 90° [
ÐAFE = ÐCFD [Vertically opposite angles]
\ DAEF ~ DCDF [By AA similarity]
(ii)
In D‘s ABD and CBE, we have
ÐABD = ÐCBE = ÐB [Common angle]
ÐADB = ÐCEB = 90° [
\ DABD ~ DCBE [By AA similarity]
(iii)
In D‘s AEF and ADB, we have
ÐAEF = ÐADB = 90° [
ÐFAE = ÐDAB [Common angle]
\ DAEF ~ DADB [By AA similarity]
(iv)
In D‘s FDC and BEC, we have
ÐFDC = ÐBEC = 90° [
ÐFCD = ÐECB [Common angle]
\ DFDC ~ DBEC [By AA similarity]
Question:
In figure, if BD ^ AC and CE ^ AB, prove that
(i) DAEC ~ DADB
(ii)
Solution:
Given: BD ^ AC and CE ^ AB
To prove:
(i) DAEC ~ DADB
(ii)
Proof:
(i)
In D‘s AEC and ADB,
we have
ÐAEC = ÐADB = 90° [
and, ÐEAC = ÐDAB [Each equal to ÐA]
\ D AEC ~ DADB [By AA similarity]
(ii)
We have,
DAEC ~ DADB [Proved above]
Þ
Þ
Question:
If CD and GH (D and H lie on AB and FE) are respectively bisectors of ÐACB and ÐEGF and DABC ~ DFEG, prove that
(i)
(ii) DDCB ~ DHGE
Solution:
Given:
(i) DABC ~ DFEG
(ii) CD, the bisector of ÐACB meets AB at D and
(iii) GH, the bisector of ÐEGF meets FE at H.
To prove:
(i)
(ii) DDCB ~ DHGE
Proof:
(i)
DABC ~ DFEG [Given]
\ ÐBAC = ÐEFG …(i)
ÐABC = ÐFEG …(ii)
and ÐACB = ÐFGE …(iii)
[Corresponding angles of similar triangles]
\
\ ÐACD = ÐFGH …(iv)
[As CD is the bisector of ÐACB]
and ÐDCB = ÐHGE …(v)
[As GH is the bisector of ÐEGF]
In DACD and DFGH, we have
ÐDAC = ÐHFG [From (i)]
ÐACD = ÐFGH [From (iv)]
DACD ~ DFGH [By AA similarity]
Þ
(ii)
In DDCB and DHGE, we have
ÐDBC = ÐHEG [From (ii)]
ÐDCB = ÐHGE [From (v)]
\ DDCB ~ DHGE [By AA similarity]