# CBSE 10th Mathematics | Similar Triangles | Criteria For Similarity Of Two Triangles

## 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 $\displaystyle A\prime B\prime C\prime$ are similar if

(i)    $\displaystyle \angle A=\angle A\prime$, $\displaystyle \angle B=\angle B\prime$, $\displaystyle \angle C=\angle C\prime$ and

(ii)    $\displaystyle \frac{AB}{A\prime B\prime }=\frac{BC}{B\prime C\prime }=\frac{CA}{C\prime A\prime }$

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 $\displaystyle \angle A=\angle D$, $\displaystyle \angle B=\angle E$, $\displaystyle \angle C=\angle F$.

To prove:

DABC ~ DDEF

Proof:

Case 1: When AB = DE

In triangles ABC and DEF, we have

$\displaystyle \angle A=\angle D$            [Given]

AB = DE             [Given]

$\displaystyle \angle B=\angle E$            [Given]

\    $\displaystyle \Delta ABC\cong \Delta DEF$        [By ASA congruency]

Þ    BC = EF and AC = DF    [c.p.c.t.]

Thus,    $\displaystyle \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$        [corresponding sides of similar Ds are proportional]

Hence, $\displaystyle \Delta ABC\tilde{\ }\Delta DEF$

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 $\displaystyle \cong$ $\displaystyle \Delta DPQ$        [By SAS congruency]

\    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

\    $\displaystyle \frac{DP}{DE}=\frac{DQ}{DF}$            [Corollary to BPT Theorem]

\    $\displaystyle \frac{AB}{DE}=\frac{AC}{DF}$            [$\displaystyle \because$ DP = AB and DQ = AC (by construction)]

Similarly $\displaystyle \frac{AB}{DE}=\frac{BC}{EF}$

\    $\displaystyle \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

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 $\displaystyle \cong$ DDEF     [By SAS congruency]

\    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

\    $\displaystyle \frac{AP}{AB}=\frac{AQ}{AC}$        [Corollary to BPT Theorem]

\    $\displaystyle \frac{DE}{AB}=\frac{DF}{AC}$        [$\displaystyle \because$ AP = DE and AQ = DF (by construction)]

Similarly,     $\displaystyle \frac{DE}{AB}=\frac{EF}{BC}$

Thus,         $\displaystyle \frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC}$

or        $\displaystyle \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

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 $\displaystyle \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

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:         $\displaystyle \frac{AB}{DE}=\frac{AC}{DF}$            [Given]

Þ     $\displaystyle \frac{DP}{DE}=\frac{DQ}{DF}$                        …(i)

[$\displaystyle \because$ DP = AB and DQ = AC (by construction)]

In DDEF, we have

$\displaystyle \frac{DP}{DE}=\frac{DQ}{DF}$            [From (i)]

\    PQ || EF             [By the converse of BPT]

\    ÐDPQ = ÐDEF and ÐDQP = ÐDFE [Corresponding angles]

\    DDPQ = ~ DDEF         [By AA similarity]    …(ii)

\    $\displaystyle \frac{DP}{DE}=\frac{PQ}{EF}$

or    $\displaystyle \frac{AB}{DE}=\frac{PQ}{EF}$            [$\displaystyle \because$ DP = AB]        …(iii)

But    $\displaystyle \frac{AB}{DE}=\frac{BC}{EF}$             [given]            …(iv)

From equations (iii) and (iv), we have $\displaystyle \frac{PQ}{EF}=\frac{BC}{EF}$

Þ    BC = PQ                         …(v)

In DABC and DDPQ, we have

AB = DP [By construction]

AC = DQ [By construction]

BC = PQ         [By (v)]

\    DABC $\displaystyle \cong$ DDPQ     [by SSS congruency]

Þ    DABC ~ DDPQ     [$\displaystyle \because$ DABC $\displaystyle \cong$ DPQ Û
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 $\displaystyle \frac{AB}{DE}=\frac{AC}{DF}$

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,     $\displaystyle \frac{AB}{DE}=\frac{AC}{DF}$                        …(ii)

\    $\displaystyle \frac{DP}{DE}=\frac{DQ}{DF}$        [$\displaystyle \because$ AB = DP and AC = DQ (by construction)]

In DDEF, we have

$\displaystyle \frac{DP}{DE}=\frac{DQ}{DF}$        [From (ii)]

\    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.

### 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

$\displaystyle \frac{PQ}{DE}=\frac{QR}{EF}=\frac{PR}{DF}=\frac{1}{2}$

Therefore, by SSS-criterion of similarity, we have

DPQR ~ DDEF

(iii)

In D‘s MNP and EFG, we observe that

$\displaystyle \frac{NP}{FG}=\frac{MP}{EG}\ne \frac{MN}{EF}$

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°    [$\displaystyle \because$ ÐN = 180° – ÐMÐP = 180° – 70° – 30° = 80°]

So, by AA-criterion of similarity DDEF ~ DMNP.

### 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:    $\displaystyle \frac{a}{p}=\frac{b}{q}=\frac{c}{r}=\frac{a+b+c}{p+q+r}$

Proof:

Since DABC and DPQR are similar, therefore their corresponding sides are proportional.

Let     $\displaystyle \frac{a}{p}=\frac{b}{q}=\frac{c}{r}=k$                        …(i)

Þ    a = kp, b = kq and c = kr

\    $\displaystyle \frac{\text{perimeter of }\Delta \text{ABC}}{\text{perimeter}\ \text{of }\Delta \text{PQR}}=\frac{a+b+c}{p+q+r}=\frac{kp+kq+kr}{p+q+r}=\frac{k(p+q+r)}{(p+q+r)}$

= $\displaystyle \frac{k(p+q+r)}{(p+q+r)}=k$                    … (ii)

From (i) and (ii), we get $\displaystyle \frac{a}{p}=\frac{b}{q}=\frac{c}{r}=\frac{a+b+c}{p+q+r}=\frac{\text{perimeter of}\ \Delta \text{ABC }}{\text{perimeter of }\Delta \text{PQR}}$

[each equal to k]

### 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    $\displaystyle \frac{AB}{DE}=\frac{AC}{DF}$

Þ    $\displaystyle \frac{12}{x}=\frac{8}{40}\Rightarrow \frac{12}{x}=\frac{1}{5}\Rightarrow x=60$ metres

### 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]

### 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]

### Solution:

Given: $\displaystyle \frac{QT}{PR}=\frac{QR}{QS}$ and Ð1 = Ð2

To prove:
DPQS ~ DTQR

Proof: We have,

$\displaystyle \frac{QT}{PR}=\frac{QR}{QS}$    [Given]

Þ    $\displaystyle \frac{QT}{QR}=\frac{PR}{QS}$                            …(i)

We also have,

Ð1 = Ð2                            [Given]

Þ    PR = PQ [Sides opposite to equal angles are equal]        …(ii)

From (i) and (ii), we get

$\displaystyle \frac{QT}{QR}=\frac{PQ}{QS}$

Þ    $\displaystyle \frac{PQ}{QT}=\frac{QS}{QR}$

Thus, in triangles PQS and TQR, we have

$\displaystyle \frac{PQ}{QT}=\frac{QS}{QR}$ and $\displaystyle \angle PQS=\angle TQR=\angle Q$

\    DPQS ~ DTQR.    [By SAS similarity]

### Solution:

Given: DABE DACD

To prove:

Proof: Since DABE DACD

\    AB = AC [cpct]…. (i)

and,    AD = AE                    …. (ii)

Also,     $\displaystyle \frac{AB}{AD}=\frac{AC}{AE}$    [From (i) and (ii)]

Þ    $\displaystyle \frac{AB}{AC}=\frac{AD}{AE}$                    … (iii)

Thus, in triangles ADE and ABC, we have

$\displaystyle \frac{AB}{AC}=\frac{AD}{AE}$

and,    ÐBAC = ÐDAE                [Common]

Hence, by SAS criterion of similarity.

### 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°            [$\displaystyle \because$ CE AB and AD BC]

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°            [$\displaystyle \because$ AD BC and CE AB]

\    DABD ~ DCBE                [By AA similarity]

(iii)

In D‘s AEF and ADB, we have

ÐAEF = ÐADB = 90°            [$\displaystyle \because$ AD BC and CE AB]

ÐFAE = ÐDAB            [Common angle]

\    DAEF ~ DADB                [By AA similarity]

(iv)

In D‘s FDC and BEC, we have

ÐFDC = ÐBEC = 90°            [$\displaystyle \because$ AD BC and CE AB]

ÐFCD = ÐECB            [Common angle]

\    DFDC ~ DBEC                [By AA similarity]

### Solution:

Given: BD AC and CE AB

To prove:

(i)    DAEC ~ DADB

(ii)    $\displaystyle \frac{CA}{AB}=\frac{CE}{DB}$

Proof:

(i)

In D‘s AEC and ADB,

we have

ÐAEC = ÐADB = 90°                [$\displaystyle \because$ CE AB and BD AC]

and,    ÐEAC = ÐDAB                [Each equal to ÐA]

\    AEC ~ DADB                [By AA similarity]

(ii)

We have,

DAEC ~ DADB                [Proved above]

Þ    $\displaystyle \frac{CA}{BA}=\frac{EC}{DB}$

Þ    $\displaystyle \frac{CA}{AB}=\frac{CE}{DB}$

### 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)    $\displaystyle \frac{CD}{GH}=\frac{AC}{FG}$

(ii)    DDCB ~ DHGE

Proof:

(i)

DABC ~ DFEG            [Given]

\    ÐBAC = ÐEFG            …(i)

ÐABC = ÐFEG            …(ii)

and    ÐACB = ÐFGE            …(iii)

[Corresponding angles of similar triangles]

\    $\displaystyle \frac{1}{2}\angle ACB=\frac{1}{2}\angle FGE$

\    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    $\displaystyle \frac{CD}{GH}=\frac{AC}{FG}$

(ii)

In DDCB and DHGE, we have

ÐDBC = ÐHEG            [From (ii)]

ÐDCB = ÐHGE            [From (v)]

\    DDCB ~ DHGE            [By AA similarity]

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