CHAPTER- 3 UNIT- 3

CONGRUENCY OF TRIANGLES

EXERCISE 3.3.3

1. In a △ ABC, AB =AC and ∟A = 𝟓𝟎𝟎. Find ∟B and ∟C.

Ans: ∟B =∟C (AB = AC) A

∟B = ∟C = (AB = AC) 50°

∟B = ∟C = 180 °−50°

2

= 130

2 = 650

∴ ∟B = 650 and ∟C = 650

2. In △ABC , AB=BC and ∟B= 𝟔𝟒𝟎. Find ∟C.

Ans: ∟A = ∟C (AB = BC ) B

∟A = ∟C = 180 °−64°

2 640

= 116 °

2 =580

∟A = 580 and ∟C = 580 A C

∴ ∟C = 580

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

a) ∟B = ∟C (AB = AC)

∟B = ∟C = 180 °−40°

2 =

140 °

2 = 700

∴ ∟B = 700 and ∟C = 700

∟C + x = 1800 (Linear pair)

X = 1800 − 700

x= 1100

b) A

65

B C D

AC=CD

∟A = ∟D = 300 (Therorem 5)

In △ ACD ∟A + ∟C + ∟D = 1800

(Sum of all angles of △ in 1800)

300 + ∟C + 300 = 1800

∟C = 1800 − 600

∟C = 1200

∟ACB + ∟ACD = 1800 (Linear pair)

∟ACB = 1800 − 1200 = 600

In △ ABC,∟A + ∟B + ∟C = 1800

(Sum of all angles of △ is 1800 )

650 + x+ 600 = 1800

X = 1800 − 1250

X= 550

C)

30°

55°

x

A

B C

75°

Ans: ∟D = ∟C = 55° (AB=AC)

∟ADB + ∟ADC = 180° (Linear Pair)

∟ADC = 180° - 75° = 105°

In ∆ADC, ∟A + ∟D + ∟C =180°

(sum of all angles of ∆ is 180°)

A= 105° + 55° = 180°

∟A = 180° - 160°

∟A = 20°

∴ x = 20°

d)

Ans: ∟B = 50° (Theorem 5)

In ∆ADB, ∟A +∟B + ∟D = 180° (sum of all angles of ∆ 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 ∆ ACD, ∟A = ∟C (AD = AC)

∟A = ∟C = 180 °−100 °

2=

80 °

2 = 40°

∴ X = 40°

4) Suppose ABC is an equilateral triangle. Its base BC is produced to D such

that BC= CD. Calculate (i) ∟ACD and (ii) ∟ADC.

Data: ABC is an equilateral triangle

AB = BC = CA

BC = CD.

Proof:

In ∆ ABC, ∟A = ∟B = ∟C = 60°

(Equilateral Triangle)

∟ACB = 60°

∟ACB +∟ACD =180° (linear pair)

∟ACD = 180° - 60°

∟ACD = 120°

In ∆ 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.

Ans: 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 ∆ BEC are ∆ CDB,

1) BC = BC (common side)

2) ∟ABC = ∟ACB (AB = AC)

3) ∟E = ∟D (90°)

∴ BEC ≅ ∆ CBD (ASA postulate)

∴ BE = DC (congruency property)

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

bisects BC.

Ans: Data: AD bisects BC, AD ⊥ B

BD = DC

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

Proof: In ∆ ADB and ∆ ADC

1) AD = AD (common side)

2) BD = DC (Data)

3) ∟ADB ∟ADC = (90°)

∴ ∆ ADB ≅ ∆ ADC (sas postulate)

∴ AB = AC (congruency property)

∴ ∆ ABC is an Isosceles triangle.

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

Ans: Data: AB = BC = CA ,

∆ ABC is an equilateral triangle.

To Prove: ∟A = ∟B = ∟C

Proof: In ∆ ABC,

1) ∟A = ∟B (AC = BC)

2) ∟B = ∟C (AB = AC)

3) ∟C = ∟A (BA = BC)

∴ ∆ ABC is also equiangular.

EXERCISE 3.3.4

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

also the midpoint of AC.

Ans: Data: AB | | DC

P is the midpoint of BD, DP = PB.

To Prove: „P‟ is the midpoint of AC .

Proof: In ∆ ABF and ∆ PDC,

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

2) ∟DPC = ∟PAB (Vertically opposite angles) 3) DP = PB (Data)

∴ ∆ ABP ≅ ∆ 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.

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

AC=AB A

To Prove: AE = AD.

Prove: In ΔABE and ΔACD, D E

1) ∟BEA =∟CDA (90°) B C 2) AB =AC (Data) 3) ∟A = ∟A (Common)

∴ ∆ ABE ≅ ∆ ACD (ASA Postulate)

∴ AE = AD (Congrency 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.

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

To Prive: „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 (Congrency property)

∴ PO = OQ (Congrency 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.

Ans: Data : 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 (Congrency property)

5. Suppose ABC is an equiangular triangle. Prove that it is equilateral. (You

have seen earier that an equilateral triangle is equiangular. Thus for

triangles triangularity is equivalent to equilaterality)

Ans : 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)

EXERCISE 3.3.5

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

Δ ADC ≅ Δ ADB.

Ans : Data: AC = AB A

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) B D C

∴ ΔADC ≅ ADB (SSS postulate)

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

≅ ΔQOR ≅ ΔROS ≅ ΔSOP.

Ans: 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?

Ans: A P

B D C Q S R

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.

Data: In Δ PQR, PQ = QR, ∟P = ∟R

L, M and N are the mid points of

The sides of PQ, QR and RP respectively.

To Prove: LN = MN Q

Proof: In Δ LNP and Δ NMR,

1. ∟P = ∟R (PQ = QR) L M

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

3. PL = RM (L and M are the P R

Midpoints of the sides PQ and QR respectively)

∴ Δ LNP ≅ Δ MNR (SSS postulate)

∴ LN = MN (Congruency propert

EXERCISE 3.3.6

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

ADC are congruent.

Ans:

A B

D C

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.

Ans: Data: ABC is a triangle, A

D is the midpoint if BC.

DE ⊥ AB, DF ⊥ AC, DF = DF E F

To Prove: Δ ABC is an isosceles triangle,

AB = AC ∟B =∟C B D 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.

Data: Abc is a triangle, A

BE ⊥ AC and CF ⊥ AB,

BE = CF F E

To Prove: Δ ABC is an isosceles triangle,

AB = AC B C

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)

∴ Δ ABC is an isosceles triangle.

EXERCISE 3.3.7

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

sides.

Ans: 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.

Ans: 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.

Ans: 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)

Ans: Data: ABCD is a quadrilateral.

AD is the largest side and B C

BC is the smallest side.

To prove: ∟A < ∟C

Constuction: Join AC.

Proof: AD is the largest side is ΔACD. A D

∴ 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).

Ans: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, A

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 B C

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

ADDITIONAL PROBLEMS ON “CONGRUENCY OF TRIANGLES”

1. Fill in the blanks to make the statements true.

a) In right triangle the hypotenuse is the ……….. side. (longest)

b) The sum of three altitudes of a triangle is ……… than is perimeter. (less)

c) The sum of any two sides of a triangle is ………. Than the third side. (greater)

d) If two angles of a triangle are unequal, then the smaller angle has the ….. side

opposite to it. (smaller)

e)Difference of any two sides of triangle is …….. than the third side. (less)

f) If two sides of a triangle are unequal, then the larger side has …….. angle

opposite to it. (larger)

2. Justify the following statement with reasons:

a) The sum of three sides of a triangle is more than the sum of its altitudes.

In ΔABC, Let AD , BE , CF

Be the altitudes.

We have to prove that:

AB + BC + AC > AD , BE , CF .

Consider ΔADC, ∟ADC = 90°,

∴ AC → hypotenure we know that hypotenure is the longest side of a Rt. Led Δ.

∴ AC > AD ……………(1)

Similarly : AB > AE , BC > BF ……(2)

Compare AB + BC + AC & AD , BE , CF

AB + BC + AC > AD, BE , CF

[from (1) and (2)]

B) The sum of any two sides of a triangle is greater than twice the median

drawn to the third side. A

In Δ ABC,AD in the median.

T.P.T : AB + AC > 2 AD

Solution:

We know that B D C

AB + AC > BC [Sum of any two sides

and a triangle in Greater

than its third side]

But ΔABD In ΔADC

AB + BD > AD

AD < AB and AD <AC

In ΔABD and ADC

AD is common

BD = DC

∴ ∟A = ∟A ∴AB + BD > AD

AC + DC > AD

∴ AB > AD and

AC > AD

Adding AB +AC > 2 AD

c) Difference of any two sides of triangle is less than the third side.

In Δ ABC, we should A

Prove that: AB – AC <2 BC

B C

From the statement, sum of any two sides of a triangle is greater than the third side.

AB + BC > AC

[Changing the inequality]

3. Two triangle ABC and DBC have common base BC. Suppose AB=DC and ∟ABC = ∟BCD . Prove that AC = BD.

Given: Δ ABC and DBC stand on same base BC.

AB = DC; ∟ABC = ∟BCD A D

T.P.T : AC = BD

Proof : Consider ∆𝑠 ABC and BDC

BC = BC (given or common base) B C

AB = DC (given) and ∟ABC = ∟BCD (given)

By SAS postulate ΔABC ≅ ΔBDC

∴ Corresponding sides of congruent ∆𝑠 are equal.

∴ AC = BD

4. Let AB and CD be two line segments such that AD and BC intersect at O. Suppose AO=OC and BO = OD. Prove that AB = CD.

Given: AD intersect BC at O.

AO = OC BO = OD

T.P.T : AB = CD

Proof : Consider ∆𝑠 AOB and COD .

∟O = ∟O [Vertically opposite angles]

AO = OC [given]

∴ Δ AOB ≅ Δ COD [SAS postulate]

∴ AB = CD [Corresponding sides of congruent ∆𝑠 ]

5. Let ABC be a triangle. Draw a triangle BDC externally on BC such that

AB = BD. Prove that and AC = CD. Prove that Δ ABC ≅ ΔDBC.

Given: ∆𝑠 ABC and DBC A

AB = BD

AC = DC

T.P.T: ΔABC ≅ DBC B C

Proof : In ∆𝑠 ABC and BDC

AB = BD

AC = DC (given) D

BC is common side

∴ Δ ABC ≅ Δ DBC [SSS postulate]

6. Let ABCD be a sqare and let points P on AB and Q on DC be such that

DP=AQ. Prove that BP = CQ. D Q C

Given: ABCD is a square

DP = AQ

T.P.T. : BP = CQ A P B

Proof: Consider ∆𝑠 APD and AQD

AD is common side

DP = AQ (given)

∟PAD = ∟ADQ =90° [ABCD is a square]

∴ According RHS postulate

Δ DPA ≅ ΔAQD

⇒ AP = QD

But AB = CD [sides of a square]

∴ AB – AP = CD – QD

⇒ BP = CQ (Axiom)

7. In a triangle ABC, AB = AC. Suppose P is point on AB and Q is a point on AC

such that a PΔAP€ ≅ ΔAQB.

Given: ΔABC, AB = AC A

AP = AQ

T.P.T.: Δ APC and Δ AQB are congruent. P Q

Proof: In Δ APC and Δ AQB

B C

AP = AQ (given )

AC = AB (given)

∟A is common

∴ ΔAPC ≅ Δ AQB (SAS Postulate)

8. In an isoscelesbtriangle , if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

Let Δ ABC, be isosceles A

AB = AC

Vertex ∟Le = ∟BAC

∟BAC = 2 (∟ABC + ∟ACB) B C

Solution : AB = AC

⇒ ∟B = ∟C = x

∴ In Δ ABC, ∟A +∟B + ∟C =180°

[ Angle sum of Δ in 180°]

∴ ∟A + ∟B + ∟C = 180°

2 (∟ABC + ∟ACB) + x + x =180°

2(x +x) + x + x = 180°

4x + 2x = 180°

6x = 180°

X = 180

6 = 30°

∴ ∟A = 2(30 + 30) = 120°

∟B = ∟C = 30°

9. If the bisector of the vertical angle of a triangle bisects the base, show that the triangle is isosceles.

Given : In ΔABC, AD bisects ∟BAC and A

BD = DC

T.P.T. : AB = AC

Proof: In ∆𝑠 ABD and ADC

BD = DC (given) B D C

AD is common

∟BAD = ∟CAD (given)

∴ ΔABD ≅ Δ ADC (SAS Postulate)

⇒ AB = AC [Corresponding sides of congruent ∆𝑠 ]

10. Suppose ABC is an isosceles triangle with AB = AC . Side BA is produced to D such that BA = AD. Prove that ∟BCD is a right angle.

Given: In ΔABC AB = AC D

AB = AD

T.P.T.: ∟BCD = 90°

Proof : In ABC, AB = AC A

∴ ∟ABC = ∟ACB (Base les) C

IIIly in ΔADC B C

AC = AD (given)

∴ ∟ACD = ∟ACB = x and

∟ACD = ∟ADC = y ∟ADC = ∟BDC

In Δ BDC, ∟BDC + ∟DCB + ∟CBD = 180°

∟BDC + ∟DCA + ∟ACB + ∟CBD = 180°

y+y+x+x = 180

2y + 2x = 180

X + y = 90°

i.e., ∟BCA + ∟ACD =90° = ∟BCD

11. Suppose ABC is a triangle and D is the midpoint of BC. Suppose further that the perpendiculars from D to AB and AC are equal. Prove that the triangle is

isosceles.

Given: In ABC ,D is midpoint of BC A

∴ BD = DC, ∟E = ∟F = 90°, DE = DF

T.P.T. : ΔABC is isosceles AB = AC E F

Proof: In ∆s BDE and DFC

∟E = ∟F = 90° B C

BD = DC

ED =DF [given]

∴ ΔBDE ≅ Δ DFC

⇒ BE = CF and also ∟B = ∟C

If ∟B = ∟C then AC = AB

∴ Δ ABC is isosceles

12. Let AB,CD be two line segments such that AB║CD and AD║ BC. Let E be the

midpoint of BC and let DE extended meet AB in F. Prove that AB =BF.

Given: AB ║ CD , AD ║BC, A B F

E→ Midpoint of

BC – BE = EC E

T.P.T : AB = BF D C

Proof : In ∆s DCE and BFE

CE = BE = [given]

∟EDC = ∟BFC [alternate ∟les]

∟DCB = ∟CBF

∴ ΔDCE ≅ Δ BFC [SAS Postulate]

⇒ DC = BF

But DC = AB

∴ BF = AB A