PDF of RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence are provided here. Students who aim to secure good marks in their board marks can refer to RD Sharma Solutions. These solutions are formulated by the BYJUâ€™S expert team in Maths to help students solve the problems.

Chapter 16, Congruence includes five exercises. Students can easily access answers to the problems present in RD Sharma Solutions for Class 7. Let us have a look at some of the important concepts that are being discussed in this chapter.

- Definition of congruent figures
- Congruence of line segments
- Congruence of two angles
- Congruence of two squares
- Congruence of two rectangles
- Congruence of two circles
- Congruence of two triangles
- Sufficient conditions for Congruence of two triangles
- The side-side-side(SSS) Congruence condition
- SAS Congruence condition
- ASA Congruence condition
- RHS Congruence condition

## Download the PDF of RD Sharma Solutions For Class 7 Maths Chapter 16 Congruence

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 16 – Congruence

Exercise 16.1 Page No: 16.3

**1. Explain the concept of congruence of figures with the help of certain examples.**

**Solution:**

Congruent objects or figures are exact copies of each other or we can say mirror images of each other. The relation of two objects being congruent is called congruence.

Consider Ball 1 and Ball 2. These two balls are congruent.

Ball 1 Ball 2

Now consider the two stars below. Star A and Star B are exactly the same in size, colour and shape. These are congruent stars

Star A Star B

**2. Fill in the blanks:**

**(i) Two line segments are congruent if â€¦â€¦..**

**(ii) Two angles are congruent if â€¦â€¦..**

**(iii) Two square are congruent if â€¦â€¦â€¦**

**(iv) Two rectangles are congruent if â€¦â€¦â€¦**

**(v) Two circles are congruent if â€¦â€¦.**

**Solution:**

(i) They are of equal lengths

(ii) Their measures are the same or equal.

(iii) Their sides are equal or they have the same side length

(iv) Their dimensions are same that is lengths are equal and their breadths are also equal.

(v) They have same radii

**3. In Fig. 6,Â âˆ POQÂ â‰…âˆ ROS, can we say thatÂ âˆ PORÂ â‰…âˆ QOS**

**Solution:**

Given that

âˆ POQÂ â‰…âˆ ROSÂ

Also given that âˆ ROQÂ â‰…âˆ ROQÂ

Therefore addingÂ âˆ ROQÂ toÂ bothÂ sidesÂ of âˆ POQÂ â‰…âˆ ROS,Â

We get, âˆ POQÂ + âˆ ROQÂ â‰…âˆ ROQÂ + âˆ ROSÂ

Therefore, âˆ PQRÂ â‰…âˆ QOS

**4. In fig. 7, a = b = c, name the angle which is congruent toÂ âˆ AOC**

**Solution:**

From the figure we have

âˆ Â AOB =Â âˆ Â BOC =Â âˆ Â COD

Therefore,Â âˆ Â AOB =Â âˆ COD

Also,Â âˆ Â AOB +Â âˆ Â BOC =Â âˆ Â BOC +Â âˆ Â COD

âˆ Â AOC =Â âˆ Â BOD

Hence,Â âˆ Â BODÂ â‰…âˆ Â AOC

**5. Is it correct to say that any two right angles are congruent? Give reasons to justify your answer.**

**Solution:**

Two right angles are congruent to each other because they both measure 90^{o}.

We know that two angles are congruent if they have the same measure.

**6. In fig. 8,Â âˆ AOCÂ â‰…âˆ PYRÂ andÂ âˆ BOCÂ â‰…âˆ QYR. Name the angle which is congruent toÂ âˆ AOB.**

**Solution:**

Given thatâˆ AOC â‰…âˆ PYR

Also given that âˆ BOCÂ â‰…âˆ QYRÂ

Now, âˆ AOCÂ = âˆ AOBÂ + âˆ BOCÂ âˆ PYRÂ =âˆ PYQÂ +âˆ QYRÂ

ByÂ puttingÂ theÂ valueÂ ofÂ âˆ AOCÂ andÂ âˆ PYRÂ inÂ âˆ AOC â‰…âˆ PYR

WeÂ get, âˆ AOBÂ + âˆ BOCÂ â‰…âˆ PYQÂ + âˆ QYRÂ âˆ AOBÂ â‰…âˆ PYQÂ (âˆ BOCÂ â‰…âˆ QYR)Â

Hence, âˆ AOBÂ â‰…âˆ PYQ

**7. Which of the following statements are true and which are false;**

**(i) All squares are congruent.**

**(ii) If two squares have equal areas, they are congruent.**

**(iii) If two rectangles have equal areas, they are congruent.**

**(iv) If two triangles have equal areas, they are congruent.**

**Solution:**

(i) False.

**Explanation:**

All the sides of a square are of equal length. However, different squares can have sides of different lengths. Hence all squares are not congruent.

(ii) True.

**Explanation:**

Two squares that have the same area will have sides of the same lengths. Hence they will be congruent.

(iii) False

**Explanation:**

Area of a rectangle = length x breadth

Two rectangles can have the same area. However, the lengths of their sides can vary and hence they are not congruent.

(iv) False

**Explanation:**

Area of a triangle = 12 x base x height

Two triangles can have the same area but the lengths of their sides can vary and hence they cannot be congruent.

Exercise 16.2 Page No: 16.8

**1. In the following pairs of triangle (Fig. 12 to 15), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.**

**Solution:**

(i) InÂ Î”Â ABC andÂ Î”Â DEF

AB = DE = 4.5 cm (Side)

BC = EF = 6 cm (Side) and

AC = DF = 4 cm (Side)

SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.

Therefore, by SSS criterion of congruence,Â Î”ABCÂ â‰…Â Î”DEF

(ii) InÂ Î”Â ACB andÂ Î”Â ADB

AC = AD = 5.5cm (Side)

BC = BD = 5cm (Side) and

AB = AB = 6cm (Side)

SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.

Therefore, by SSS criterion of congruence,Â Î”ACBÂ â‰…Â Î”ADB

(iii) InÂ Î”Â ABD andÂ Î”Â FEC,

AB = FE = 5cm (Side)

AD = FC = 10.5cm (Side)

BD = CE = 7cm (Side)

SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.

Therefore, by SSS criterion of congruence,Â Î”ABDÂ â‰…Â Î”FEC

(iv) InÂ Î”Â ABO andÂ Î”Â DOC,

AB = DC = 4cm (Side)

AO = OC = 2cm (Side)

BO = OD = 3.5cm (Side)

Therefore, by SSS criterion of congruence,Â Î”ABOÂ â‰…Â Î”ODC

**2. In fig.16, AD = DC and AB = BC**

**(i) IsÂ Î”ABDÂ â‰…Â Î”CBD?**

**(ii) State the three parts of matching pairs you have used to answer (i).**

**Solution:**

(i) YesÂ Î”ABD â‰…Î”CBD by the SSS criterion.

Hence Î”ABD â‰…Î”CBD

(ii) We have used the three conditions in the SSS criterion as follows:

AD = DC

AB = BC and

DB = BD

**3. In Fig. 17, AB = DC and BC = AD.**

**(i) IsÂ Î”ABCÂ â‰…Â Î”CDA?**

**(ii) What congruence condition have you used?**

**(iii) You have used some fact, not given in the question, what is that?**

**Solution:**

(i) From the figure we have AB = DC

BC = AD

And AC = AC

Therefore by SSS criterionÂ Î”ABCÂ â‰…Â Î”CDA

(ii) We have used Side congruence condition with one side common in both the triangles.

(iii)Yes, have used the fact that AC = CA.

**4. InÂ Î”PQRÂ â‰…Â Î”EFD,**

**(i) Which side ofÂ Î”PQRÂ equals ED?**

**(ii) Which angle ofÂ Î”PQRÂ equals angle E?**

**Solution:**

(i) PR = ED

Since the corresponding sides of congruent triangles are equal.

(ii)Â âˆ QPR =Â âˆ FED

Since the corresponding angles of congruent triangles are equal.

**5. Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?**

**It âˆ B = 50Â°, what is the measure of âˆ R?**

**Solution:**

Given that AB = AC in isoscelesÂ Î”ABC

And PQ = PR in isoscelesÂ Î”PQR.

Also given that AB = PQ and QR = BC.

Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)

Hence,Â Î”ABCÂ â‰…Â Î”PQR

Now

âˆ ABC =Â âˆ PQR (Since triangles are congruent)

However,Â Î”PQR is isosceles.

Therefore,Â âˆ PRQ =Â âˆ PQR =Â âˆ ABC = 50^{o}

**6. ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If âˆ BAC = 40Â° andÂ âˆ BDC = 100Â°, then findÂ âˆ ADB.**

**Solution:**

Given ABC and DBC are both isosceles triangles on a common base BC

âˆ BAD =Â âˆ CAD (corresponding parts of congruent triangles)

âˆ BAD +Â âˆ CAD = 40^{o}/ 2Â

âˆ BAD = 40^{o}

âˆ BAD = 40^{o}/2 =20^{o}

âˆ ABC +Â âˆ BCA +Â âˆ BAC = 180^{o} (Angle sum property)

SinceÂ Î”ABC is an isosceles triangle,

âˆ ABC =Â âˆ BCAÂ

âˆ ABC +âˆ ABC + 40^{o }= 180^{o}

2Â âˆ ABC = 180^{o}â€“ 40^{o} = 140^{o}

âˆ ABC = 140^{o}/2 = 70^{o}

âˆ DBC +Â âˆ Â BCD +Â âˆ Â BDC = 180^{o} (Angle sum property)

SinceÂ Î”ABC is an isosceles triangle,Â âˆ Â DBC =Â âˆ BCDÂ

âˆ DBC +Â âˆ DBC + 100^{oÂ }= 180^{o}

2Â âˆ DBC = 180Â°â€“ 100^{o}Â = 80^{o}

âˆ DBC = 80^{o}/2 = 40^{o}

InÂ Î”Â BAD,

âˆ ABD +Â âˆ BAD +Â âˆ ADB = 180^{o }(Angle sum property)

30^{o} + 20^{o} +Â âˆ ADB = 180^{o} (âˆ ADB =Â âˆ ABC â€“Â âˆ DBC),Â

âˆ ADB = 180^{o}– 20^{o}â€“ 30^{o}

âˆ ADB = 130^{o}

âˆ ADB =130^{o}

**7. Î”Â ABC andÂ Î”ABD are on a common base AB, and AC = BD and BC = AD as shown in Fig. 18. Which of the following statements is true?**

**(i)Â Î”ABCÂ â‰…Â Î”ABD**

**(ii)Â Î”ABCÂ â‰…Â Î”ADB**

**(iii)Â Î”ABCÂ â‰…Â Î”BAD**

**Solution:**

InÂ Î”ABC andÂ Î”BAD we have,

AC = BD (given)

BC = AD (given)

And AB = BA (corresponding parts of congruent triangles)

Therefore by SSS criterion of congruency, Î”ABCÂ â‰…Â Î”BAD

Therefore option (iii) is true.

**8. In Fig. 19,Â Î”ABC is isosceles with AB = AC, D is the mid-point of base BC.**

**(i) IsÂ Î”ADBÂ â‰…Â Î”ADC?**

**(ii) State the three pairs of matching parts you use to arrive at your answer.**

**Solution:**

(i) Given that AB = AC.

Also since D is the midpoint of BC, BD = DC

Also, AD = DA

Therefore by SSS condition,

Î”ADBÂ â‰…Â Î”ADC

(ii)We have used AB, AC; BD, DC and AD, DA

**9. In fig. 20,Â Î”ABC is isosceles with AB = AC. State ifÂ Î”ABCÂ â‰…Â Î”ACB. If yes, state three relations that you use to arrive at your answer.**

**Solution:**

Given that Î”ABC is isosceles with AB = AC

Î”ABCÂ â‰…Â Î”ACBby SSS condition.

Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB

**10. Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? DoesÂ âˆ ABD equalÂ âˆ ACD? Why or why not?**

**Solution:**

Yes, congruent because given that ABC and DBC have side BC common, AB = BD and AC = CD

Also from the above data we can say

By SSS criterion of congruency,Â Î”ABCÂ â‰…Â Î”DBC

No,Â âˆ ABD andÂ âˆ ACD are not equal because ABÂ not equal toÂ AC

Exercise 16.3 Page No: 16.14

**1. By applying SAS congruence condition, state which of the following pairs (Fig. 28) of triangle are congruent. State the result in symbolic form**

**Solution:**

(i) From the figure we have OA = OC and OB = OD and

âˆ AOB =Â âˆ COD which are vertically opposite angles.

Therefore by SAS condition, Î”AOC â‰…Î”BOD

(ii) From the figure we have BD = DC

âˆ ADB =Â âˆ ADC = 90^{o} and

Therefore, by SAS condition, Î”ADB â‰…Î”ADC.

(iii) From the figure we have AB = DC

âˆ ABD =Â âˆ CDB and

Therefore, by SAS condition, Î”ABD â‰…Î”CBD

(iv) We have BC = QR

ABC = PQR = 90^{o}

And AB = PQ

Therefore, by SAS condition,Â Î”ABCâ‰…Â Î”PQR.

**2. State the condition by which the following pairs of triangles are congruent.**

**Solution: **

(i) AB = AD

BC = CD and AC = CA

Therefore by SSS condition, Î”ABCâ‰… Î”ADC

(ii) AC = BD

AD = BC and AB = BA

Therefore, by SSS condition, Î”ABD â‰… Î”ADC

(iii) AB = AD

âˆ BAC =Â âˆ DAC and

Therefore by SAS condition, Î”BAC â‰… Î”BAC

(iv) AD = BC

âˆ DAC =Â âˆ BCA and

Therefore, by SAS condition, Î”ABC â‰… Î”ADC

**3.** **In fig. 30, line segments AB and CD bisect each other at O. Which of the following statements is true?**

**(i)Â Î”AOC â‰… Î”DOB**

**(ii)Â Î”AOC â‰… Î”BOD**

**(iii)Â Î”AOC â‰… Î”ODB**

**State the three pairs of matching parts, you have used to arrive at the answer.**

**Solution:**

From the figure we have,

And, CO = OD

Also, AOC = BOD

Therefore, by SAS condition,Â Î”AOC â‰… Î”BOD

**4. Line-segments AB and CD bisect each other at O. AC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles that are given or otherwise known. Are the two triangles congruent? State in symbolic form, which congruence condition do you use?**

**Solution:**

We have AO = OB and CO = OD

Since AB and CD bisect each other at 0.

AlsoÂ âˆ AOC =Â âˆ BOD

Since they are opposite angles on the same vertex.

Therefore by SAS congruence condition, Î”AOC â‰… Î”BOD

**5. Î”ABC is isosceles with AB = AC. Line segment AD bisectsÂ âˆ A and meets the base BC in D.**

**(i) Is Î”ADB â‰… Î”ADC?**

**(ii) State the three pairs of matching parts used to answer (i).**

**(iii) Is it true to say that BD = DC?**

**Solution:**

(i) We have AB = AC (Given)

âˆ BAD =Â âˆ CAD (AD bisectsÂ âˆ BAC)

Therefore by SAS condition of congruence, Î”ABD â‰… Î”ACD

(ii) We have used AB, AC;Â âˆ BAD =Â âˆ CAD; AD, DA.

(iii) Now, Î”ABDâ‰…Î”ACD

Therefore by corresponding parts of congruent triangles

BD = DC.

**6. In Fig. 31, AB = AD and âˆ BAC = âˆ DAC.**

**(i) State in symbolic form the congruence of two triangles ABC and ADC that is true.**

**(ii) Complete each of the following, so as to make it true:**

**(a) âˆ ABC =**

**(b) âˆ ACD =**

**(c) Line segment AC bisects â€¦.. And â€¦â€¦..**

**Solution:**

i) AB = AD (given)

âˆ BAC = âˆ DAC (given)

AC = CA (common)

Therefore by SAS condition of congruency, Î”ABC â‰… Î”ADC

ii) âˆ ABC = âˆ ADC (corresponding parts of congruent triangles)

âˆ ACD = âˆ ACB (corresponding parts of congruent triangles)

**7. In fig. 32, AB || DC and AB = DC.**

**(i) Is Î”ACD â‰… Î”CAB?**

**(ii) State the three pairs of matching parts used to answer (i).**

**(iii) Which angle is equal to âˆ CAD?**

**(iv) Does it follow from (iii) that AD || BC?**

**Solution:**

(i) Yes by SAS condition of congruency, Î”DCA â‰… Î”BAC

(ii) We have used AB = DC, AC = CA and âˆ DCA = âˆ BAC.

(iii) âˆ CAD = âˆ ACB since the two triangles are congruent.

(iv) Yes this follows from AD parallel to BC as alternate angles are equal. lf alternate angles are equal the lines are parallel

Exercise 16.4 Page No: 16.19

**1. Which of the following pairs of triangle are congruent by ASA condition?**

** **

**Solution:**

(i) We have,

Since âˆ ABO = âˆ CDO = 45^{o} and both are alternate angles, AB parallel to DC, âˆ BAO = âˆ DCO (alternate angle, AB parallel to CD and AC is a transversal line)

âˆ ABO = âˆ CDO = 45^{o} (given in the figure) Also,

AB = DC (Given in the figure)

Therefore, by ASA Î”AOB â‰… Î”DOC

(ii) In ABC,

Now AB =AC (Given)

âˆ ABD = âˆ ACD = 40^{o} (Angles opposite to equal sides)

âˆ ABD + âˆ ACD + âˆ BAC = 180^{o} (Angle sum property)

40^{o} + 40^{o} + âˆ BAC = 180^{o}

âˆ BAC =180^{o} – 80^{o} =100^{o}

âˆ BAD + âˆ DAC = âˆ BAC

âˆ BAD = âˆ BAC â€“ âˆ DAC = 100^{o} â€“ 50^{o} = 50^{o}

âˆ BAD = âˆ CAD = 50Â°

Therefore, by ASA, Î”ABD â‰… Î”ADC

(iii) In Î” ABC,

âˆ A + âˆ B + âˆ C = 180^{o }(Angle sum property)

âˆ C = 180^{o}– âˆ A â€“ âˆ B

âˆ C = 180^{o} â€“ 30^{o} â€“ 90^{o} = 60^{o}

In PQR,

âˆ P + âˆ Q + âˆ R = 180^{o} (Angle sum property)

âˆ P = 180^{o} â€“ âˆ Q â€“ âˆ R

âˆ P = 180^{o }â€“ 60^{o }â€“ 90^{o} = 30^{o}

âˆ BAC = âˆ QPR = 30^{o}

âˆ BCA = âˆ PRQ = 60^{o} and AC = PR (Given)

Therefore, by ASA, Î”ABC â‰… Î”PQR

(iv) We have only

BC = QR but none of the angles of Î”ABC and Î”PQR are equal.

Therefore, Î”ABC and Cong Î”PRQ

**2. In fig. 37, AD bisects A and AD and AD âŠ¥ BC.**

**(i) Is Î”ADB â‰… Î”ADC?**

**(ii) State the three pairs of matching parts you have used in (i)**

**(iii) Is it true to say that BD = DC?**

**Solution:**

(i)Â Yes, Î”ADBâ‰…Î”ADC, by ASA criterion of congruency.

(ii) We have used âˆ BAD = âˆ CAD âˆ ADB = âˆ ADC = 90^{o}

Since, AD âŠ¥ BC and AD = DA

(iii) Yes, BD = DC since, Î”ADB â‰… Î”ADC

Â

**3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.**

**Solution:**

We have drawn

Î” ABC with âˆ ABC = 65^{o} and âˆ ACB = 70^{o}

We now construct Î”PQR â‰… Î”ABC has âˆ PQR = 65^{o} and âˆ PRQ = 70^{o}

Also we construct Î”PQR such that BC = QR

Therefore by ASA the two triangles are congruent

**4. In Î” ABC, it is known that âˆ B = C. Imagine you have another copy of Î” ABC**

**(i) Is Î”ABC â‰… Î”ACB**

**(ii) State the three pairs of matching parts you have used to answer (i).**

**(iii) Is it true to say that AB = AC?**

**Solution: **

(i) Yes Î”ABC â‰… Î”ACB

(ii) We have used âˆ ABC = âˆ ACB and âˆ ACB = âˆ ABC again.

Also BC = CB

(iii) Yes it is true to say that AB = AC since âˆ ABC = âˆ ACB.

**5. In Fig. 38, AX bisects âˆ BAC as well as âˆ BDC. State the three facts needed to ensure that Î”ACD â‰… Î”ABD **

**Solution:**

As per the given conditions,

âˆ CAD = âˆ BAD and âˆ CDA = âˆ BDA (because AX bisects âˆ BAC)

AD = DA (common)

Therefore, by ASA, Î”ACD â‰… Î”ABD

**6. In Fig. 39, AO = OB and âˆ A = âˆ B.**

**(i) Is Î”AOC â‰… Î”BOD**

**(ii) State the matching pair you have used, which is not given in the question.**

**(iii) Is it true to say that âˆ ACO = âˆ BDO?**

**Solution:**

We have

âˆ OAC = âˆ OBD,

AO = OB

Also, âˆ AOC = âˆ BOD (Opposite angles on same vertex)

Therefore, by ASA Î”AOC â‰… Î”BOD

Exercise 16.5 Page No: 16.23

**1. In each of the following pairs of right triangles, the measures of some part are indicated alongside. State by the application of RHS congruence conditions which are congruent, and also state each result in symbolic form. (Fig. 46)**

**Solution:**

(i) âˆ ADC = âˆ BCA = 90^{o}

AD = BC and hypotenuse AB = hypotenuse AB

Therefore, by RHS Î”ADB â‰… Î”ACB

(ii) AD = AD (Common)

Hypotenuse AC = hypotenuse AB (Given)

âˆ ADB + âˆ ADC = 180^{o} (Linear pair)

âˆ ADB + 90^{o} = 180^{o}

âˆ ADB = 180^{o} â€“ 90^{o} = 90^{o}

âˆ ADB = âˆ ADC = 90^{o}

Therefore, by RHS Î” ADB = Î” ADC

(iii) Hypotenuse AO = hypotenuse DO

BO = CO

âˆ B = âˆ C = 90^{o}

Therefore, by RHS, Î”AOBâ‰…Î”DOC

(iv) Hypotenuse AC = Hypotenuse CA

BC = DC

âˆ ABC = âˆ ADC = 90^{o}

Therefore, by RHS, Î”ABC â‰… Î”ADC

(v) BD = DB

Hypotenuse AB = Hypotenuse BC, as per the given figure,

âˆ BDA + âˆ BDC = 180^{o}

âˆ BDA + 90^{o} = 180^{o}

âˆ BDA= 180^{o }â€“ 90^{o} = 90^{o}

âˆ BDA = âˆ BDC = 90^{o}

Therefore, by RHS, Î”ABD â‰… Î”CBD

**2. Î” ABC is isosceles with AB = AC. AD is the altitude from A on BC.**

**(i) Is Î”ABD â‰… Î”ACD?**

**(ii) State the pairs of matching parts you have used to answer (i).**

**(iii) Is it true to say that BD = DC?**

**Solution:**

(i) Yes, Î”ABD â‰… Î”ACD by RHS congruence condition.

(ii) We have used Hypotenuse AB = Hypotenuse AC

AD = DA

âˆ ADB = âˆ ADC = 90^{o} (AD âŠ¥ BC at point D)

(iii)Yes, it is true to say that BD = DC (corresponding parts of congruent triangles)

Since we have already proved that the two triangles are congruent.

Â

**3. Î”ABC is isosceles with AB = AC. Also. AD âŠ¥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condition do you use? Which side of ADC equals BD? Which angle of Î” ADC equals âˆ B?**

**Solution:**

We have AB = AC â€¦â€¦ (i)

AD = DA (common) â€¦â€¦ (ii)

And, âˆ ADC = âˆ ADB (AD âŠ¥ BC at point D) â€¦â€¦ (iii)

Therefore, from (i), (ii) and (iii), by RHS congruence condition, Î”ABD â‰… Î”ACD, the triangles are congruent.

Therefore, BD = CD.

And âˆ ABD = âˆ ACD (corresponding parts of congruent triangles)

**4. Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.**

**Solution:**

Consider

Î” ABC with âˆ B as right angle.

We now construct another triangle on base BC, such that âˆ C is a right angle and AB = DC

Also, BC = CB

Therefore, BC = CB

Therefore by RHS, Î”ABC â‰… Î”DCB

Â

**5.In fig. 47, BD and CE are altitudes of Î” ABC and BD = CE.**

**(i) Is Î”BCD â‰… Î”CBE?**

**(ii) State the three pairs or matching parts you have used to answer (i)**

**Solution:**

(i) Yes, Î”BCD â‰… Î”CBE by RHS congruence condition.

(ii) We have used hypotenuse BC = hypotenuse CB

BD = CE (Given in question)

And âˆ BDC = âˆ CBE = 90^{o}