Congruency is one of the fundamental concepts in geometry. This concept is used to classify the geometrical figures on the basis of their shapes. Two geometrical figures are said to be congruent, if they have same shape and size. For example:

- Two line segments are congruent if they have same length.
- Two angles are congruent if they have same measures.
- Two circles are congruent if they have same radii.
- Two squares are congruent if they have sides of same measures.

**Congruency of Triangles:**

Two triangles are congruent **if all the angles and sides of one triangle are equal to the corresponding angles and sides of the other triangle.**

In triangle ABC and DEF, we observe that, AB = DE, AC = DF and BC = EF; ∠A = ∠D, ∠B = ∠E and ∠C = ∠F.

Therefore, triangles ABC and DEF are congruent. We write this as,

∆ABC ≅ ∆DEF

A few words about the use fo this notation. When we write ∆ABC ≅ ∆DEF, the important things is to observe that the vertices A, B and C corresponding to the vertices D, E and F in that order. If we write ∆ABC ≅ ∆EFD, this gives a different meaning. This means AB = EF, BC = FD and CA = DE; ∠A = ∠E, ∠B = ∠F and ∠C = ∠D.

**Corresponding sides and angles**

Let us say that, on superscription, triangle ABC covers triangle DEF exactky in such way that,

- AB = DE, AC = DF and BC = EF;
- ∠A = ∠D, ∠B = ∠E and ∠C = ∠F.

Then, ∆ABC ≅ ∆DEF

Angles which coincide on superscription are called **corresponding angles. **Sides which coincide on superscription are called **corresponding sides. **

**Exercise 3.3.1**

- Identify the corresponding sides and corresponding angles in the following congruent triangles.

Solution:

PQ = XY, PR = XZ , QR = YZ

(ii)

PR = AC, QP = AB

2. Pair of congruent triangles and incomplete statements related to them are given below. Observe the figures carefully and fill up the blanks.

(a) In the adjoining figure if ∠C = ∠F, then AB = —– and BC = ————-

(b)In the adjoining figure if BC = EF, then, ∠C =—– and ∠A = ————-

(c) In the adjoining figure if AC = CE and ∆ABC ≅ ∆DEF ∠D =——- and ∠A = ———–

Solution:

(a) AB = DE and BC = EF

(b) ∠C =∠F and ∠A = ∠D

(c) ∠D = ∠B and ∠A = ∠E

**3.3.2 SAS postulates for the congruency of triangles**

**Sides-angles-sides postulates [SAS postulates]**

**If the two sides and included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle, then the two triangles are congruent.**

In triangles ABC and DEF, we observe that AB = DE, AC = DF and ∠A = ∠D. Hence, SAS postulate tells,

∆ABC ≅ ∆DEF

**Example 1: **In the figure O is the midpoint of AB and CD. Prove that (i) ∆AOC ≅ ∆BOD

(ii) AC = BD

**Solution:**

In triangles AOC and BOD, we have,

AO = BO, (O, the midpoint of AB)

∠AOC = ∠BOD, (vertically opposite angles)

CO = OD, (O, the midpoint of CD)

So, by SAS postulate we have,

∆AOC ≅ ∆BOD

Hence, AC = BD, as they are corresponding parts of congruent triangles.

**Example 2: **In the figure, it is given that AE = AD and BD = CE. Prove that ∆AEB is congruent ∆ADC.

**Solution:**

We have AE = AD and CE = BD, adding we get AE + CE = AD + BD

Therefore, AC = AB

In triangles AEB and ADC, we have AE = AD (given)

AB = AC (proved)

∠EAB = ∠DAC (common angle)

By SAS postulate

∆AEB ≅ ∆ADC

**Example 3: **In a quadrilateral ACBD, AC = AD and AB bisect ∠A. Show that, ∆ABC is congruent to ∆ABD

**Solution:**

In triangle ABC and ABD, we have

AC = AD, (given)

∠CAB = ∠DAB , (AB bisects ∠A)

AB = AB (common side)

Hence,

∆AEB ≅ ∆ABD

**Exercise 3.3.2**

- In the following figure PQRS is a rectangle, identify the congruent triangles formed by the diagonals.

Solution:

∆POS ≅ ∆QOR

∆POQ ≅ ∆SOR

2. In the figure ABCD is a square, M,N,O and P are the midpoints of sides AB, BC, CD and DA respectively. Identify the congruent triangles.

**Solution:**

∆AMP ≅ ∆NDO ≅ ∆COP ≅ ∆APM

3. In a triangle, ABC, AB = AC. Points E on AB and D on AC are such that AE = AD. Prove that triangles BCD and CBE are congruent.

**Solution:**

We have AB = AC and Points E on AB and D on AC are such that AE = AD

adding we get AE + BE = AB and AD + DC = AC

Therefore, EB = DE

The line EC and BD coincides at a point, say O, then, ∠EOB = ∠DOC (common angle)

AB = AC (Data) ,then BO = OC.

By SAS postulate

∆BCD ≅ ∆CBE

4. In the adjoining figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that DE||BC [Hint: use the concept of alternate angles]

**Solution:**

We have AB = AD and CA = AE .

Then, ∠EAD = ∠BAC (common angle)

By SAS postulate

∆BCD ≅ ∆CBE

Therefore,

DE||BC

**3.3.3 Consequences of SAS postulate:**

**Theorem 1: In a triangle, the angles opposite to equal sides are equal.**

Given: A triangle ABC in which AB = AC

To prove: ∠C = ∠B

Construction: Draw the angle bisector of ∠A. Let it cut BC at D. Let us compare triangles ABD and ACD.

Proof:

Statement:

AB = AC (given)

AD = AD (common side)

∠BAD = ∠CAD (by construction)

We can use SAS postulate to conclude that ∆ADB ≅ ∆ADC.

Hence ∠ABC = ∠ACB, since these are corresponding angles of congruent triangles. Thus the theorem is proved.

**Example 4**: In the figure AB = AC and DB = DC. Prove that ∠ABD = ∠ACD

Solution:

In ∆ABC we have,

AB = AC

Therefore, ∠ABC = ∠ACB (angles opposite to equal sides)

Again in ∆DBC, we have DB = DC(given)

Then, ∠DBC = ∠DCB (angles opposite to equal sides)

Hence we obtain,

∠ABC – ∠DBC = ∠ACB – ∠DCB

∠ABD = ∠ACD

**exercise 3.3.3**

- In a △ ABC, AB =AC and ∟A = 𝟓𝟎°. Find ∟B and ∟C.

Solution:

∟B =∟C (Since AB = AC)

∟B + ∟C + ∟A = 180° (Since AB = AC)

∟B + ∟C = 180°−50°

2∟C = 180° – 50°

∟C = 130°/2

∴ ∟B = 65° and ∟C = 65°

2.In △ABC , AB=BC and ∟B= 𝟔𝟒°. Find ∟C.

Solution:

∟A = ∟C (Since AB = BC)

∟A + ∟C + ∟B = 180°

2∟C = 180° – 64° = 116°

∟C = 180°/2 =58°

∟A = 58° and ∟C = 58°

∴ ∟C = 58°

3.In each of the following figure, Find the value of x:

(i)

Solution:

∟B = ∟C (Since AB = BC)

∟A + ∟C + ∟B = 180°

2∟C = 180° – 40° = 140°

∟C = 140°/2 =70°

∟A = 70° and ∟C = 70°

∴ ∟C = 70°

We have ∟ACB + ∟ACD = 180°

∟ACD = x (data)

x = 180° – 70° = 110°

(ii)

Solution:

∟A = ∟D = 30° (Therorem 5)

In △ ACD ∟A + ∟C + ∟D = 180°

(Sum of all angles of △ in 180°)

30° + ∟C + 30° = 180°

∟C = 180° − 60°

∟C = 120°

∟ACB + ∟ACD = 180° (Linear pair)

∟ACB = 180°− 120°= 60°

In △ ABC,∟A + ∟B + ∟C = 180°

(Sum of all angles of △ is 180°)

65° + x+ 60° = 180°

X = 180°− 125°

X= 55°

(iii)

Solution:

∟D = ∟C = 55° (AB=AC)

∟ADB + ∟ADC = 180° (Linear Pair)

∟ADC = 180° – 75° = 105°

In ΔADC, ∟A + ∟D + ∟C =180°

(sum of all angles of Δ is 180°)

A= 105° + 55° = 180°

∟A = 180° – 160°

∟A = 20°

∴ x = 20°

(iv)

Solution:

∟B = 50° (Theorem 5)

In ΔADB, ∟A +∟B + ∟D = 180° (sum of all angles of Δ is 180°)

50° + 50° +∟D = 180°

∟D = 180° – 100° = 80°

∟BDA + ∟ADC = 180° (Linear pair)

∟ADC = 180°- 80° = 100°

∟A = ∟C = 50° (AD=DC)

In Δ ACD, ∟A = ∟C (AD = AC)

∟A = ∟C = (180°−100°)/2= 80°/2 = 40°

∴ X = 40°

4. 4) Suppose ABC is an equilateral triangle. Its base BC is produced to D such that BC= CD. Calculate (i) ∟ACD and (ii) ∟ADC.

Solution:

Data: ABC is an equilateral triangle

AB = BC = CA

BC = CD.

Proof:

In Δ ABC, ∟A = ∟B = ∟C = 60°

(Equilateral Triangle)

∟ACB = 60°

∟ACB +∟ACD =180° (linear pair)

∟ACD = 180° – 60°

∟ACD = 120°

In Δ ACD, ∟A = ∟D (AC = CD)

∟A = ∟D = (180°−120°)/2 = 60°/2 = 30°

∴ ∟ADC = 30°

5. Show that the perpendiculars drawn from the vertices of the base of an isosceles triangle to the opposite sides are equal.

Solution:

Given: ABC is an isosceles triangle, AB = AC.

BE and CD are the perpendiculars drawn from the vertices of the base B and C.

To Prove: BE = CD

Proof: In ΔBEC are ΔCDB,

1) BC = BC (common side)

2) ∟ABC = ∟ACB (AB = AC)

3) ∟E = ∟D (90°)

∴ BEC ≅ Δ CBD (ASA postulate)

∴ BE = DC (congruency property)

6) Prove that a Δ abc is an isosceles triangle if the altitude AD from A on BC bisects BC.

Solution:

Data: AD bisects BC, AD ⊥ B

BD = DC

To Prove: Δ ABC is an isosceles triangle, AB = AC.

Proof: In Δ ADB and Δ ADC

1) AD = AD (common side)

2) BD = DC (Data)

3) ∟ADB ∟ADC = (90°)

∴ Δ ADB ≅ Δ ADC (sas postulate)

∴ AB = AC (congruency property)

∴ Δ ABC is an Isosceles triangle.

7) Suppose a triangle is equilateral. Prove that it is equiangular.

Solution:

Data: AB = BC = CA ,

Δ ABC is an equilateral triangle.

To Prove: ∟A = ∟B = ∟C

Proof: In Δ ABC,

1) ∟A = ∟B (AC = BC)

2) ∟B = ∟C (AB = AC)

3) ∟C = ∟A (BA = BC)

∴ Δ ABC is also equiangular.

**3.3.4 ASA postulate for congruency**

If two angles and the included side of a triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.

Given two triangles ABC and DEF such that, ∠B = ∠E, ∠C = ∠F and BC = EF, the ASA postulate tells,

∆ABC ≅ ∆DEF

**Theorem 2: If in a triangle two angkes are equal, then the sidfes opposite to them are equal. (converse of theorem1)**

Given: Triangle ABC in which ∠B = ∠C

To prove: AC = AB

Construction: draw AD⊥BD

Proof: Then ∠ADB = ∠ADC = 90. We are given ∠DBA = ∠DCA. Consider triangles ADB and ADC. We have,

∠ADB + ∠DBA + ∠BAD = 180 = ∠ADC + ∠DCA + ∠CAD

IT follows that, ∠BAD = ∠CA

Consider triangles ADB and ADC. We have,

∠BAD = ∠CAD (proved)

∠ADB = ∠ADC (both are right angles)

AD = AD (common side)

Therefore,

∆ADB ≅ ∆ADC, by ASA condition. We conclude that, AB = AC by property of congruency, This completes the proof of the converse of the theorem 1.

**Example 5: **In a triangle ABC, AB = AC and the bisectors of angles B and C intersect at O. Prove that BO = CO and AO is the bisector of angle ∠BAC

Solution:

Since the angles opposite to equal sides are equal.

AB = AC

Therefore, ∠C = ∠B

That implies, ∠B/2 = ∠C/2

Since BO and CO are the bisectors if ∠B and ∠C, we also have

∠ABO = ∠B/2 = ∠C/2 = ∠ACO

Consider, ∆BCO;

∠OBC = ∠OCB

Thus, BO = CO

Finally, consider triangles ABO and ACO

BA = CA (given)

BO = CO (proved)

∠ABO = ∠ACO (proved)

Hence by SAS postulate,

∆ABO ≅ ∆ACO

Therefore, ∠BAO = ∠CAO

AO bisects ∠A

Example 6: Diagonal AC of a quadrilateral ABCD bisects the angles and ∠A and ∠C. Prove that AB = AD and CB = CD.

Solution:

Since the diagonal AC bisects the angles ∠A and ∠C

We have, ∠BAC = ∠DAC and ∠BCA = ∠DCA. In triangles ABC and ADC, we have,

∠BAC = ∠DAC (given)

∠BCA = ∠DCA (given)

AC = AC (common side)

So by ASA postulate, we have,

∆BAC ≅ ∆DAC

BA = AD and CB = CD

**Exercise 3.3.4**

- In the given figure, If AB || DC and P is the midpoint of BD, Prove that P is also the midpoint of AC.

Solution:

Data: AB | | DC

P is the midpoint of BD, DP = PB.

To Prove: P is the midpoint of AC .

Proof: In Δ ABF and Δ PDC,

1) ∟DCP = ∟PAB (Alternate angles, AB | | DC)

2) ∟DPC = ∟PAB (Vertically opposite angles)

3) DP = PB (Data)

∴ Δ ABP ≅ Δ PDC (ASA Postulate)

∴ AP = PC (Congruency property)

∴ P is the midpoint of AC (Congruency property)

2. In the adjacent figure, CD and BE are altitudes of an isosceles triangle ABC with AC=AB. Prove that AE=AD.

Solution:

ΔABC is an isosceles triangle with CD and BE are altitudes of it.

AC=AB

To Prove: AE = AD.

Prove: In ΔABE and ΔACD

1) ∟BEA =∟CDA (90°)

2) AB =AC (Data)

3) ∟A = ∟A (Common)

∴ Δ ABE ≅ Δ ACD (ASA Postulate)

∴ AE = AD (Congruency property)

3. In figure, AP and BQ are perpendiculars to the line segment AB and AP = BQ. Prove that O is the mid point of line segment AB as well as PQ.

Solution : Data : AP and BQ are perpendiculars to line segment AB. AP = BQ

To Prove: O is the midpoint of line

Segment AB and PQ.

Proof: In Δ APQ and Δ BOQ

1) ∟PAO = ∟OBQ (90°)

2) ∟PQA = ∟QOB (V.O.A)

∴ Δ APO ≅ Δ BOQ (ASA postulate)

∴ AO = OB (Congruency property)

∴ PO = OQ (Congruency property)

∴ O is the mid-point of line segments

AB and PQ (Congruency property)

4. Suppose ABC is an isosceles triangle with AB = AC; BC and CE are bisectors of ∟B and ∟C. Prove that BD = CE.

Solution:

ABC is an isosceles triangle with

AB = AC,∟B = ∟C. A

To prove: BD = CE

Proof in Δ ABD and Δ ACE

1) ∟BAD = ∟CAE (Common angle) E D

2) ∟ABD = ∟ACE (Data)

3) AB = AC (Data) B C

∴ Δ ABD ≅ Δ ACE (ASA postulate)

∴ BD = CE (Congruency property)

5. Suppose ABC is an equiangular triangle. Prove that it is equilateral. (You have seen earlier that an equilateral triangle is equiangular. Thus for triangles triangularity is equivalent to equilaterality)

Solution :

Data : ABC is an equiangular triangle.

∟A = ∟B = ∟C

To prove AB = BC = CA

Proof : In Δ ABC,

1) AB = BC (∟A = ∟C)

2) BC = CA (∟B =∟A)

3) CA = AB (∟C = ∟B)

∴ AB = BC = CA (From the above proof)

**3.3.5 SSS postulate for congruency**

**If three sides of one triangle are equal to the three corresponding sides of the other triangle, then the triangles are congruent.**

Example 9: In the figure, iit is given that AB = CD and AD = BC. Prove that triangles ADC and CBA are congruent.

Solution:

In triangles ADC and CBA, we have,

AB = CD (given)

AD = BC (given)

AC = AC(common side)

By SSS congruency condition, Δ ADC ≅ ΔCBA

Example 10: In the figure AD = BC and BD = CA. Prove that ∠ADB = ∠BCA and ∠DAB = ∠CBA

Solution:

AD = BC (given)

AC = BD (given)

AB = AB(common side).

We can use SSS condition to conclude that Δ ABD ≅ ΔBAC. From this we conclude that

∠ADB = ∠BCA and ∠DAB = ∠CBA

**Exercise 3.2.5**

1. In a triangle ABC, AC = AB and the altitude AD bisects BC. Prove that Δ ADC ≅ Δ ADB.

Solution :

Data: AC = AB

AD bisects BC, BD = DC

To prove: ΔADC ≅ Δ ADB

Proof : In Δ ADC and Δ ADB,

1. AB = AC (Data)

2. AD = AD (Common side)

3. BD = DC (Data)

∴ ΔADC ≅ ADB (SSS postulate)

2. In a square PQRS, diagonals bisect each other at O. prove that ΔPOQ ≅ ΔQOR ≅ ΔROS ≅ ΔSOP.

Solution:

Data: PQRS is a square, diagonals bisect each other at O,

PO = OR = SO = OQ

PQ = QR = SR = PS

To Prove: ΔPOQ ≅ ΔQOR ≅ ΔROS ≅ ΔSOP

Proof:

1) In ΔPOQ and ΔSOR

1. PO = OR (Data)

2. SO = OQ (Data) P Q

3. PQ = SR (Data)

∴ ΔPQR ≅ ΔSOR (SSS postulate)

2)In ΔPOS and ΔQOR

1. PO = OR (Data)

2. SO = OQ (Data) S R

3. PS = QR (Data)

∴ ΔPOS ≅ QOS (SSS postulate)

∴ from (1) and (2), we get,

ΔPOQ ≅ ΔQOR ≅ ΔROS ≅ ΔSOP

3. In the figure, two sides AB, BC and the median AD of Δ ABC are respectively equal to two sides PQ, OR and median PS of Δ PQR. Prove that (i) Δ ABD ≅ ΔPSQ; (ii) Δ ADC ≅ Δ PSR. Does it follow that triangle ABC and PQR are congruent?

Solution:

Data: In Δ ABC and ΔPQR,

AB = PQ

BC = QR

AD = PS

To Prove: (i) Δ ADB ≅ ΔPSQ

(ii) Δ ADC ≅ ΔPSR

Proof : (i) In ΔADB and Δ PSQ.

1. AD = PS (Data)

2. AB = PQ (Data)

3. BD = QS (BC = QR, D and S are midpoints)

∴ Δ ADB ≅ Δ PSQ (SSS postulate)

(ii) In Δ ADS and Δ PSR

1. AD = PS (Data)

2. DC = SR(BC = QR,D and S are midpoints)

∟ADB = ∟PSQ (Congruency property)

180° – ∟ADB = 180° – ∟PSQ

3. ∟ADC = ∟PSR

∴ Δ ADC ≅ Δ PSR (SSS postulate)

In Δ ABC and Δ PQS,

1. AB = PQ (Data)

2. BC = QR (Data)

3. AC = PR (Congruency property)

∴ Δ ABC ≅ Δ PQR (SSS postulate)

4. In ΔPQR, PQ = QR; L, M and N are the midpoints of the sides of PQ, QR and RP respectively. Prove that LN = MN.

Solution:

Data: In Δ PQR, PQ = QR, ∟P = ∟R

L, M and N are the midpoints of

The sides of PQ, QR and RP respectively.

To Prove: LN = MN

Proof: In Δ LNP and Δ NMR,

1. ∟P = ∟R (PQ = QR)

2. PN = NR (N is the midpoint of RP)

3. PL = RM (L and M are the midpoints of the sides PQ and QR respectively)

∴ Δ LNP ≅ Δ MNR (SSS postulate)

∴ LN = MN (Congruency property)

**3.3.6 RHS Theorem**

Theorem 3: Two right angles are congruent if the hypotenuse and a side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle. (RHS Theorem)

Given: Two right angled triangles ABC and DEF such that,

(i) ∠B = ∠E = 90°

(ii) Hypotenuse AC = Hypotenuse DF; and

(iii) AB = DE

To prove: ΔABC ≅ ΔDEF

Construction: Produce EF to G so that EG = BC, join DG.

Proof: In triangles ABC and DEG, observe that,

AB = DE (given)

BC = EG (by construction)

∠ABC = ∠DEG (each equal to 90°)

Hence by SAS, ΔABC ≅ ΔDEG this implies, ∠ACB = ∠DGE and AC = DG. But AC = DF, by the given hypothesis. We thus get,

DG = AC = DF

In triangle DGF, we have got DG = DF(we have proved it)

This implies that, ∠G = ∠F

In triangles DEF and DEG,

∠G = ∠F (proved)

∠DEG = ∠DEF (both are equal to 90°)

Hence,

∠GDE = 180°-(∠G+∠DEG) = 180° – (∠F + ∠DEF) = ∠FDE

Consider triangles DEG and DEF, we get

DG = DF (proved)

DE = DE (common)

∠GDE = ∠FDE (proved)

Hence by SAS condition,

ΔDEG≅ ΔDEF

But we have already proved that, ΔABC ≅ ΔDEG

It follows that, ΔABC ≅ ΔDEF

Example 12: Suppose ABC is an isosceles triangle such that AB = AC and AD is the altitude from A on BC. Prove that (i) AD bisects ∠A ,(ii) AD bisects BC

Solution:

We have to show that ∠BAD = ∠CAD and BD = DC

In right angled triangles ADB and ADC, we have

AB = AC (given)

AD = AD (common side)

So by RHS congruency of triangles; we have ΔABD ≅ ΔACD. Hence ∠BAD = ∠CAD and BD = DC.

**Exercise 3.3.6**

1. Suppose ABCD is rectangle, using RHS theorem, prove that triangle ABC and ADC are congruent.

Solution:

Data: AB = DC, AD = BC

To Prove: Using RHS theorem,

ΔABC ≅ Δ ADC

Proof: In Δ ABC and ΔADC,

1. ∟B = ∟D (90°)

2. AC = AC(Common side)

3. AB = DC (Opposite sides are equal/data)

∴ Δ ABC ≅ Δ ADC (RHS postulate)

2. Suppose ABC is a triangle and D is the midpoint of BC. Assume that the perpendiculars from D to AB and AC are of equal length. Prove that ABC is isosceles.

Solution:

Data: ABC is a triangle, D is the midpoint if BC.

DE ⊥ AB, DF ⊥ AC, DF = DE

To Prove: Δ ABC is an isosceles triangle,

AB = AC, ∟B =∟C

Proof: In Δ BED and ΔDFC,

1. BD = DC (Data)

2. DE = FD (Data)

3. ∟E = ∟F (90°)

∴ Δ BED ≅ Δ DFC (ASA postulate)

∴ ∟B = ∟C (Congruency property)

∴ AB = AC (Converse of theorem 5)

∴ Δ ABC is an isosceles triangle.

3. Suppose ABC is a triangle in which BE and CF are respectively the perpendiculars to the sides AC and AB. If BE = CF, prove that triangle ABC is isosceles.

Solution:

Data: Abc is a triangle,

BE ⊥ AC and CF ⊥ AB,

BE = CF

To Prove: Δ ABC is an isosceles triangle,

AB = AC

Proof: In Δ ABE and Δ ACF,

1. ∟BAE = ∟CAF (Common angle)

2. BE = CF (Data)

3. ∟AEB = ∟AFC (90°)

∴ Δ ABE ≅ Δ ACF (ASA postulate)

∴ AB = AC (Congruency property)

∴ ΔABC is an isosceles triangle.

**3.3.7 Some consequences**

**Proposition 1: Suppose two sides of a triangle are not equal. Then the angle opposite to a larger side is greater than the angle opposite to the smaller side.**

Given: A triangle ABC in which AC > AB.

To prove: ∠B > ∠C

Construction: take apoint D on AC such that AB = AD. (this is possible since AC > AB)

Proof: In triangle ABD, we have AB = AD(by construction)

Therefore ∠ABD = ∠ADB (angles opposite to equal sides)

Now, ∠BDC is an exterior angle for triangle BCD. Hence it is larger than interior angle ∠BCD. We thus get,

∠C > ∠BDA = ∠ABD < ∠ABC = ∠B

**Proposition 2: In a triangle, if two angles are unequal, then the side opposite to the larger angle is greater than the side opposite to the smaller angle.**

Given: A triangle ABC in which ∠B > ∠C.

To prove: AC > AB

Proof: Observe that, ∠B > ∠C Therefore, AC ≠ AB.

For, AC = AB implies that ∠B = ∠C

Thus either,

AC < AB or AC > AB

IF AC > AB, then by **Proposition 1: Suppose two sides of a triangle are not equal. Then the angle opposite to a larger side is greater than the angle opposite to the smaller side. **∠B > ∠C, but; this contradicts the given hypothesis. The only possibility left out is AC > AB.

**Proposition 3: In a triangle, the sum of any two sides is greater than the third side.**

Given: A triangle ABC

To prove: AB + AC > BC

Construction: Extend BA to D such that AD = AC and join DC.

Proof:

Then, BD = BA + AD = BA + AC

Since AD = AC, we have,

∠ADC = ∠ACD(angle opposite to equal side)

Hence we obtain, ∠BCD > ∠ACD = ∠ADC = ∠BDC

In triangle BCD, we have,

∠BCD > ∠ACD this implies BD > BC (by proposition 2)

But BD = BA + AC as we have observed earlier. WE thus get

BA + AC > BC

We can similarly prove CA < AB + BC and AB < BC + CA.

**Exercise 3.3.7**

1. In a triangle ABC, ∟B=28° and ∟C =56°. Find the largest and the smallest sides.

Solution:

In Δ ABC, ∟A + ∟B + ∟C = 180°(Sum of all the angles of a Δ is 180°)

∟A =180°-(28° + 56°)

∟A = 96°

∴ Largest side = BC (∟A = 96°)

∴ Smallest side = AC (∟B = 28°)

2. In a triangle ABC, we have AB=4cm, BC=5.6 cm and CA = 7.6 cm. Write the angles of the triangles in ascending order of measures.

Solution: AC = 7.6 cm.

BC = 5.6 cm.

AB = 4 cm.

Descending Order: ∟B > ∟A > ∟C

Ascending Order: ∟C < ∟A < ∟B

3. Let ABC be a triangle such that ∟B = 70° and ∟C = 40°. Suppose D is a point on BC such that AB = AD. Prove that AB >CD.

Solution:

In Δ ADB, ∟B =∟D =70°

In ADB, ∟A + ∟D + ∟B = 180° (Sum of all the angles of Δ is 180°)

∟A = 180° – 140°

∠A= 40°

In Δ ABC, ∟A + ∟B + ∟C =180°

∟A = 180° – 110° = 70°

∟CAD + ∟DAB = 70° (Linear pair)

∟CAD = 70° – 40°

∟CAD = 30°

∴∟C = 40° → AB

∴ ∟CAD = 30° → CD

∴AB > CD

4. Let ABCD be a quadrilateral in which AD is the largest side and BC is the smallest side. Prove that ∟A < ∟C. (Hint: Join AC)

Solution:

Data: ABCD is a quadrilateral.

AD is the largest side and BC is the smallest side.

To prove: ∟A < ∟C

Construction: Join AC.

Proof: AD is the largest side is ΔACD.

∴ ∟ACD > ∟CAD………….. (1)

BC is the smallest side in Δ BAC

∴ ∟ACB > ∟CAB……………(2)

Adding (1) and (2)

∴ ∟ACD + ∟ACB > ∟CAD + ∟CAB

∴ ∟C > ∟A

5. Let ABC be a triangle and P be an interior point. Prove that AB +BC +CA <2 (PA + PB + PC).

Solution:

Data : ABC is a triangle and P is an interior point.

To prove : AB + BC + CA < 2 (PA + PB + PC)

To prove : In Δ ABC,

BC < PB + PC………..(1)

AC < PA + PC……….(2)

AB < PA + PB……….(3)

Adding (1) ,(2) and (3),

BC+AC+AB < PB + PC + PA + PC + PA + PB

AB+AC+BC< 2PB + 2PA +2PC

AB +AC+BC < 2 (PA+PB+PC)

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