## 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 are similar if

(i) , , and

(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:
**

D*ABC* ~ D*DEF*

**Proof:
**

**Case 1: **When *AB* = *DE*

In triangles *ABC* and *DEF*, we have

[Given]

*AB* = *DE* [Given]

[Given]

\ [By ASA congruency]

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

Thus, [corresponding sides of similar Ds are proportional]

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 D*ABC* and D*DPQ*, we have

*AB* = *DP* [By construction]

Ð*A* = Ð*D *[Given]

*AC* =*DQ* [By construction]

\ D*ABC* [By SAS congruency]

\ Ð*ABC* = Ð*DPQ *[c.p.c.t.]* *…(i)

But Ð*ABC *= Ð*DEF *[Given]* *…(ii)

\ Ð*DPQ* = Ð*DEF* [c.p.c.t.]

But Ð*DPQ* and Ð*DEF* are corresponding angles.

Þ *PQ* || *EF*

\ [Corollary to BPT Theorem]

\ [ *DP* = *AB* and *DQ* = *AC* (by construction)]

Similarly

\

Hence, D*ABC* ~ D*DEF*.

**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 D*APQ* and D*DEF*, we have

*AP* = *DE* [By construction]

*AQ* = *DF* [By construction]

Ð*A* = Ð*D* [Given]

\ D*APQ* D*DEF* [By SAS congruency]

\ Ð*APQ* = Ð*DEF* [c.p.c.t.] …(i)

But Ð*DEF* = Ð*ABC* [Given] …(ii)

From (i) and (ii) we have

\ Ð*APQ* = Ð*ABC*

But Ð*APQ* and Ð*ABC* are corresponding angles

\ *PQ* || *BC*

\ [Corollary to BPT Theorem]

\ [ *AP* = *DE* and *AQ* = *DF* (by construction)]

Similarly,

Thus,

or

Hence, D*ABC* ~ D*DEF*.

### 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:** D*ABC ~ *D*DEF
*

**Construction:** Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ

**Proof: ** [Given]

Þ …(i)

[ *DP* = *AB* and *DQ* = *AC* (by construction)]

In D*DEF, *we have

[From (i)]

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

\ Ð*DPQ* = Ð*DEF* and Ð*DQP* = Ð*DFE* [Corresponding angles]

\ D*DPQ = ~ *D*DEF* [By AA similarity] …(ii)

\

or [ *DP* = *AB*] …(iii)

But [given] …(iv)

From equations (iii) and (iv), we have

Þ *BC* = *PQ* …(v)

In D*ABC* and D*DPQ,* we have

*AB* = *DP* [By construction]

*AC* = *DQ* [By construction]

*BC* = *PQ * [By (v)]

\ D*ABC* DDPQ [by SSS congruency]

Þ DABC ~ DDPQ [ DABC DPQ Û

DABC ~ DDPQ] … (vi)

From equation (ii) and (vi), we get

D*ABC* ~ D*DEF*

### 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 Ð*A* = Ð*D* and

**To prove:**

D*ABC ~ *D*DEF
*

**Construction: **Let *P* and *Q* be points on *DE* and *DF* respectively such that *DP* = *AB* and *DQ* = *AC*. Join *PQ.*

**Proof: **In D*ABC* and D*DPQ*, we have

*AB *= *DP* [By construction]

*AC* = *DQ* [By construction]

Ð*A *= Ð*D* [Given]

\ D*ABC *@

D*DPQ* [By SAS congruency] …(i)

Now, …(ii)

\ [ *AB* = *DP* and *AC* = *DQ* (by construction)]

In D*DEF,* we have

[From (ii)]

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

\ Ð*DPQ* = Ð*DEF* and Ð*DQP* = Ð*DFE* [Corresponding angles]

\ D*DPQ ~ *D*DEF* [By AA similarity] …(iii)

From equations (i) and (iii), we get

D*ABC* ~*
*D

*DEF.*

### 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

Ð*A* = Ð*Q* = 40°, Ð*B* = Ð*P* = 60° and Ð*C* = Ð*R* = 80°

Therefore, by *AAA*-criterion of similarity

D*BAC *~ D*PQR *

**(ii)
**

In triangle *PQR *and *DEF*, we observe that

Therefore, by SSS-criterion of similarity, we have

D*PQR* ~ D*DEF*

**(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

Ð*D* = Ð*M* = 70°

Ð*E* = Ð*N* = 80° [ Ð*N* = 180° – Ð*M* – Ð*P* = 180° – 70° – 30° = 80°]

So, by *AA*-criterion of similarity D*DEF *~ D*MNP*.

### Question:

#### Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

### Solution:

**Given:
**

D*ABC* and D*PQR* 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 …(i)

Þ a = kp, b = kq and c = kr

\

= … (ii)

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 D*ABC *and D*DEF*, we have

Ð*A* = Ð*D* = 90° and Ð*C* = Ð*F* [Angle of elevation of the sun]

Therefore, by *AA*-criterion of similarity

D*ABC* ~ D*DEF*

Þ

Þ metres

### 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 D*ABD* ~ *ECF*.

### Solution:

**Given:
**

An isosceles D*ABC *in which *AB *= *AC*, *E *is a point on *CB *produced, *AD *^ *BC *and *EF *^ *AC*.

**To prove:
**

D*ABD* ~ D*ECF*

**Proof:**

In D*ABC*, since *AB *=* AC*,

therefore Ð*C* = Ð*B* [Ðs opposite to equal sides are equal]

In D*ABD *and D*ECF*

*
*Ð

*B*= Ð

*C*[Proved above]

and Ð*EFC *= Ð*ADB* [each 90°]

\ D*ABD *~ *ECF* [*AA *similarity]

### Question:

*E* is a point on side *AD* produced of a parallelogram *ABCD* and *BE* intersects *CD* at *F*. Prove that D*ABE* ~ D*CFB*.

### Solution:

**Given:
**

*E* is a point on side *AD* produced of a parallelogram *ABCD* and *BE* intersects *CD* at *F*

**To prove:**

D*ABE* ~ D*CFB*

**Proof: **In D‘s *ABE *and *CFB*, we have

ÐAEB = ÐCBF [Alternate angles]

ÐA = ÐC [Opposite angles of a parallelogram]

\ D*ABE* ~ D*CFB*. [By AA similarity]

### Question:

#### In figure, if and Ð1 = Ð2. Prove that D*PQS *~ D*TQR*.

### Solution:

**Given: ** and Ð1 = Ð2

**To prove:**

D*PQS *~ D*TQR*

**Proof:** We have,

[Given]

Þ …(i)

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

and

\ D*PQS *~ D*TQR*. [By SAS similarity]

### Question:

#### In figure, if D*ABE *@ D*ACD*, prove that D*ADE *~ D*ABC*.

### Solution:

**Given: **D*ABE *@ D*ACD
*

**To prove:**

D*ADE *~ D*ABC*

**Proof: **Since D*ABE *@ D*ACD
*

\ *AB* = *AC *[cpct]…. (i)

and, *AD* = *AE* …. (ii)

Also, [From (i) and (ii)]

Þ … (iii)

Thus, in triangles *ADE *and *ABC*, we have

and, Ð*BAC *= Ð*DAE* [Common]

Hence, by *SAS *criterion of similarity.

D*ADE *~ D*ABC*

### Question:

#### In figure, *AD* and *CE* are two altitudes of D*ABC*. Prove that

#### (i) D*AEF* ~ D*CDF*

#### (ii) D*ABD *~ D*CBE*

#### (iii) D*AEF* ~ D*ADB*

#### (iv) D*FDC* ~ D*BEC*

### Solution:

**Given: ***AD* and *CE* are two altitudes of D*ABC
*

**To prove:
**

(i) D*AEF* ~ D*CDF*

(ii) D*ABD *~ D*CBE*

(iii) D*AEF* ~ D*ADB*

(iv) D*FDC* ~ D*BEC*

**Proof:
**

**(i)
**

In triangles *AEF *and *CDF*, we have

ÐAEF = ÐCDF = 90° [ *CE *^ *AB* and *AD *^ *BC*]

ÐAFE = ÐCFD [Vertically opposite angles]

\ D*AEF *~ D*CDF *[By AA similarity]

**(ii)
**

In D‘s *ABD* and *CBE*, we have

Ð*ABD *= Ð*CBE *= ÐB [Common angle]

Ð*ADB *= Ð*CEB *= 90° [ *AD *^ *BC *and *CE *^ *AB*]

\ D*ABD* ~ D*CBE* [By AA similarity]

**(iii)
**

In D‘s *AEF* and *ADB*, we have

Ð*AEF *= Ð*ADB *= 90° [ *AD *^ *BC *and *CE *^ *AB*]

Ð*FAE *= Ð*DAB* [Common angle]

\ D*AEF* ~ D*ADB * [By AA similarity]

**(iv)
**

In D‘s *FDC* and *BEC*, we have

Ð*FDC *= Ð*BEC *= 90° [ *AD *^ *BC *and *CE *^ *AB*]

Ð*FCD *= Ð*ECB* [Common angle]

\ D*FDC* ~ D*BEC* [By AA similarity]

### Question:

#### In figure, if *BD *^ *AC *and *CE *^ *AB*, prove that

#### (i) D*AEC* ~ D*ADB*

#### (ii)

### Solution:

**Given: ***BD *^ *AC *and *CE *^ *AB*

**To prove:
**

(i) D*AEC* ~ D*ADB*

(ii)

**Proof:
**

**(i)
**

In D‘s *AEC *and *ADB*,

we have

Ð*AEC* = Ð*ADB *= 90° [ *CE *^ *AB *and *BD *^ *AC*]

and, Ð*EAC* = Ð*DAB* [Each equal to Ð*A*]

\ D *AEC* ~ D*ADB *[By AA similarity]

**(ii)
**

We have,

D*AEC *~ D*ADB* [Proved above]

Þ

Þ

### Question:

#### If *CD *and *GH *(*D *and *H *lie on *AB *and *FE*) are respectively bisectors of Ð*ACB *and Ð*EGF *and D*ABC *~ D*FEG*, prove that

#### (i)

#### (ii) D*DCB* ~ D*HGE*

### Solution:

**Given:
**

(i) D*ABC* ~ D*FEG*

(ii) *CD*, the bisector of Ð*ACB *meets *AB *at *D* and

(iii) *GH*, the bisector of Ð*EGF *meets *FE* at *H*.

**To prove:
**

(i)

(ii) D*DCB* ~ D*HGE
*

**Proof:**

**(i)
**

D*ABC* ~ D*FEG* [Given]

\ Ð*BAC* = Ð*EFG* …(i)

Ð*ABC* = Ð*FEG* …(ii)

and Ð*ACB* = Ð*FGE* …(iii)

[Corresponding angles of similar triangles]

\

\ Ð*ACD* = Ð*FGH* …(iv)

[As *CD* is the bisector of Ð*ACB*]

and Ð*DCB* = Ð*HGE* …(v)

[As *GH* is the bisector of Ð*EGF*]

In D*ACD* and D*FGH*, we have

Ð*DAC* = Ð*HFG* [From (i)]

Ð*ACD* = Ð*FGH* [From (iv)]

D*ACD *~ D*FGH* [By *AA *similarity]

Þ

**(ii)
**

In D*DCB* and D*HGE*, we have

Ð*DBC* = Ð*HEG* [From (ii)]

Ð*DCB* = Ð*HGE* [From (v)]

\ D*DCB *~ D*HGE* [By AA similarity]