# Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles]

The Selina solutions for the questions given in chapter 9, Triangles, of the Class 9 Selina textbooks are available here. In this chapter students learn about the topic of Triangles, mainly focussing on Congruency in Triangles, in detail. Solving all the questions present in the Selina textbook will help the students in scoring full marks in the exam.

The Class 9 Selina solutions maths are very easy to understand. These solutions cover all the exercise questions included in the book and are in accordance to the syllabus prescribed by the ICSE or CISCE. Here, the PDF of the Class 9 Maths chapter 9 Selina solutions is available which can be downloaded as well as viewed online. Students can also avail these Selina solutions and download it for free to practice them offline as well.

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Exercise 9A PAGE:103

1. Which of the following pairs of triangles are congruent? In each case, state the condition of congruency:

(a) In âˆ† ABC and âˆ† DEF, AB = DE, BC = EF and âˆ B = âˆ E.

(b) In âˆ† ABC and âˆ† DEF, âˆ B = âˆ E = 90o; AC = DF and BC = EF.

(c) In âˆ† ABC and âˆ† QRP, AB = QR, âˆ B = âˆ R and âˆ C = âˆ P.

(d) âˆ† PQR, AB = PQ, AC = PR and BC = QR.

(e) In âˆ† ABC and âˆ† PQR, BC = QR, âˆ A = 90o, âˆ C = âˆ R = 40o and âˆ Q=50o.

Solution:

(a) In âˆ† ABC and âˆ† DEF

AB = DE (data)

BC = EF and âˆ B = âˆ E (given)

By SAS criteria of congruency and given data we can conclude that,

âˆ† ABC and âˆ† DEF are congruent to each other.

Therefore, âˆ† ABC â‰… âˆ† DEF.

(b) Given in âˆ† ABC and âˆ† DEF,

âˆ B = âˆ E = 90o;

AC = DF

That is hypotenuse AC = hypotenuse DF

and BC = EF

By right angle hypotenuse side postulate of congruency,

The given triangles âˆ† ABC and âˆ† DEF are congruent to each other.

Therefore, âˆ† ABC â‰… âˆ† DEF.

(c) In âˆ† ABC and âˆ† QRP,

Data: AB = QR,

âˆ B = âˆ R and

âˆ C = âˆ P.

By using SAS postulate and given data we can conclude that

The given triangles âˆ† ABC and âˆ† QRP are congruent to each other.

Therefore, âˆ† ABC â‰… âˆ† QRP.

(d) In In âˆ† ABC and âˆ† PQR,

Data: AB = PQ, AC = PR and BC = QR.

By using SSS postulate of congruency and given data we can conclude that

The given triangles âˆ† ABC and âˆ† PQR are congruent to each other.

Therefore, âˆ† ABC â‰… âˆ† PQR.

(e) In âˆ† ABC and âˆ† PQR,

Data: BC = QR, âˆ A = 90o,

âˆ C = âˆ R = 40o and âˆ Q=50o

But we know that sum of angle of triangle = 180o

Therefore, âˆ P + âˆ Q + âˆ R = 180o

âˆ P + 50o + 40o = 180o

âˆ P + 90o = 180o

âˆ P = 180o â€“ 90o

âˆ P = 90o

In âˆ† ABC and âˆ† PQR,

âˆ A = âˆ P

âˆ C = âˆ R

BC = QR

By ASA postulate of congruency,

The given triangles âˆ† ABC and âˆ† PQR are congruent to each other.

Therefore, âˆ† ABC â‰… âˆ† PQR.

2. The given figure shows a circle with centre O. P is mid-point of chord AB.

Show that OP is perpendicular to AB.

Solution:

Data: in the given figure O centre of circle.

P is mid-point of chord AB. AB is a chord

P is a point on AB such that AP = PB

Now we have to prove that OPÂ âŠ¥ AB

Construction: Join OA and OB

Proof: In âˆ† OAP and âˆ† OBP

OA = OB (because radii of common circle)

OP = OP (common)

AP = PB (data)

By SSS postulate of congruent triangles

The given triangles âˆ† OAP and âˆ† OBP are congruent to each other.

Therefore, âˆ† OAP â‰… âˆ† OBP.

The corresponding parts of the congruent triangles are congruent.

âˆ OPA = âˆ OPB (by Corresponding parts of Congruent triangles)

But âˆ OPA + âˆ OPB = 180o (linear pair)

âˆ OPA = âˆ OPB = 90o

Hence OP âŠ¥ AB

3. The following figure shows a circle with centre O.

If OP is perpendicular to AB, prove that AP=BP.

Solution:

Given: In the figure, O is the centre of the circle,

And AB is chord. P is a midpoint on AB such that AP = PB

We need to prove that AP = BP,

Construction: Join OA and OB

Proof: In right angle triangles âˆ† OAP and âˆ† OBP

Hypotenuse OA = Hypotenuse OB (because radii of common circle)

Side OP = OP (common)

AP = PB (data)

By SSS postulate of congruent triangles

The given triangles âˆ† OAP and âˆ† OBP are congruent to each other.

Therefore, âˆ† OAP â‰… âˆ† OBP.

The corresponding parts of the congruent triangles are congruent.

AP = BP (by Corresponding parts of Congruent triangles)

Hence the proof.

4. In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that: (i) âˆ† ABD and âˆ† ECD are congruent.

(ii) AB=EC

(iii) AB is parallel to EC.

Solution:

Given âˆ† ABC in which D is the mid-point of BC

We need to prove that

(i) âˆ† ABD â‰… âˆ† ECD

(ii) AB = EC

(iii) AB âˆ¥ EC

(i) In âˆ† ABD and âˆ† ECD

BD = DC (D is the midpoint of BC)

âˆ ADB = âˆ CDE (vertically opposite angles)

By SAS postulate of congruency of triangles, we have

âˆ† ABD â‰… âˆ† ECD

(ii) The corresponding parts of congruent triangles are congruent

Therefore, AB = EC (corresponding parts of congruent triangles)

(iii) Also, we have âˆ DAB = âˆ DEC (corresponding parts of congruent triangles)

AB âˆ¥ EC [âˆ DAB = âˆ DEC are alternate angles]

5. A triangle ABC has âˆ B = âˆ C. Prove that:

(i) The perpendiculars from the mid-point of BC to AB and AC are equal.

(ii) The perpendiculars form B and C to the opposite sides are equal.

Solution:

(i) Given âˆ† ABC in which âˆ B = âˆ C.

DL is perpendicular from D to AB

DM is the perpendicular from D to AC.

We need to prove that

DL = DM

Proof:

In âˆ† ABC and âˆ† DMC (DL perpendicular to AB and DM perpendicular to AC)

âˆ DLB = âˆ DMC = 90o

âˆ B = âˆ C (Given)

BD = DC (D is midpoint of BC)

By AAS postulate of congruent triangles

âˆ† DLB â‰… âˆ† DMC

The corresponding parts of the congruent triangles are congruent

Therefore DL = DM

(ii) Given âˆ† ABC in which âˆ B = âˆ C.

BP is perpendicular from D to AC

CQ is the perpendicular from C to AB.

We need to prove that

BP = CQ

Proof:

In âˆ† BPC and âˆ† CQB (BP perpendicular to AC and CQ perpendicular to AB)

âˆ BPC = âˆ CQB = 90o

âˆ B = âˆ C (Given)

BC = BC (common)

By AAS postulate of congruent triangles

âˆ† BPC â‰… âˆ† CQB

The corresponding parts of the congruent triangles are congruent

Therefore BP = CQ

6. The perpendicular bisector of the sides of a triangle AB meet at I. Prove that: IA = IB = IC

Solution:

Given triangle ABC in which AD is the perpendicular bisector of BC

BE is the perpendicular bisector of CA

CF is the perpendicular bisector of AB

AD, BE and CF meet at I

We need to prove that

IA = IB = IC

Proof:

In âˆ† BID and âˆ† CID

BD = DC (given)

âˆ BDI = âˆ CDI = 90o (AD is perpendicular bisector of BC)

BC = BC (common)

By SAS postulate of congruent triangles

âˆ† BID â‰… âˆ† CID

The corresponding parts of the congruent triangles are congruent

Therefore IB = IC

Similarly, In âˆ† CIE and âˆ† AIE

CE = AE (given)

âˆ CEI = âˆ AEI = 90o (AD is perpendicular bisector of BC)

IE = IE (common)

By SAS postulate of congruent triangles

âˆ† CIE â‰… âˆ† AIE

The corresponding parts of the congruent triangles are congruent

Therefore IC = IA

Thus, IA = IB = IC

7. A line segment AB is bisected at point P and through point P another line segment PQ, which is perpendicular to AB, is drawn. Show that: QA = QB.

Solution:

Given triangle ABC in which AB is bisected at P

PQ is the perpendicular to AB

We need to prove that

QA = QB

Proof:

In âˆ† APQ and âˆ† BPQ

AP = PB (P is the midpoint of AB)

âˆ APQ = âˆ BPQ = 90o (PQ is perpendicular to AB)

BC = BC (common)

By SAS postulate of congruent triangles

âˆ† APQ â‰… âˆ† BPQ

The corresponding parts of the congruent triangles are congruent

Therefore QA = QB

8. If AP bisects angle BAC and M is any point on AP, prove that the perpendiculars drawn from M to AB and AC are equal.

Solution:

From M, draw ML such that ML is perpendicular to AB and MN is perpendicular to AC

In âˆ† ALM and âˆ† ANM

âˆ LAM = âˆ MAN (AP is the bisector of âˆ BAC)

âˆ ALM = âˆ ANM = 90o (ML is perpendicular to AB and MN is perpendicular to AC)

AM = AM (common)

By AAS postulate of congruent triangles

âˆ† ALM â‰… âˆ† ANM

The corresponding parts of the congruent triangles are congruent

Therefore ML = MN

Hence the proof.

9. From the given diagram, in which ABCD is a parallelogram, ABL is al line segment and E is mid-point of BC.

Prove that:

(i)Â  âˆ† DCEÂ â‰… âˆ†Â LBE

(ii) AB = BL.

(iii) AL = 2DC

Solution:

Given ABCD is a parallelogram in which E is the midpoint of BC

We need to prove that

(i)Â  âˆ† DCEÂ â‰… âˆ†Â LBE

(ii) AB = BL.

(iii) AL = 2DC

(i) In âˆ† DCE andÂ âˆ†Â LBE

âˆ DCE = âˆ LBE (DC parallel to AB, alternate angles)

CE = EB (E is the midpoint of BC)

âˆ DCE = âˆ LBE (vertically opposite angles)

By ASA postulate of congruent triangles

âˆ† DCE â‰… âˆ† LBE

The corresponding parts of the congruent triangles are congruent

Therefore DC = LBâ€¦. (i)

(ii) DC = ABâ€¦â€¦ (ii)

From (i) and (ii) AB = BLâ€¦â€¦ (iii)

(iii) AL = AB + BLâ€¦â€¦â€¦ (iv)

From (iii) and (iv) AL = AB + AB

AL = 2AB

AL = 2DC from (ii)

10. In the given figure, AB = DB and Ac = DC.

IfÂ âˆ ABD = 58o,

âˆ  DBC = (2x – 4)o,

âˆ ACB = y + 15oÂ and

âˆ DCB = 63o; find the values of x and y.

Solution:

Given: In the given figure, AB = DB and Ac = DC.

IfÂ âˆ ABD = 58o,

âˆ  DBC = (2x – 4)o,

âˆ ACB = y + 15oÂ and

âˆ DCB = 63o;

We need to find the values of x and y.

In âˆ† ABC andÂ âˆ†Â DBC

AB = DB (given)

AC = DC (given)

BC = BC (common)

By SSS postulate of congruent triangles

âˆ† ABC â‰… âˆ† DBC

The corresponding parts of the congruent triangles are congruent

Therefore

âˆ ACB = âˆ DCB

yo + 15o = 63o

yo = 63o â€“ 15o

yo = 48o

âˆ ACB = âˆ DCB (corresponding parts of the congruent triangles)

But âˆ DCB = (2x â€“ 4)o

We have âˆ ACB + âˆ DCB = âˆ ABD

(2x â€“ 4)o + (2x â€“ 4)o = 58o

4x â€“ 8o = 58o

4x = 58o + 8o

4x = 66o

x = 66o/4

x = 16.5o

Thus, the values of x and y are

x = 16.5o and y = 48o

Exercise 9B PAGE:125

1. On the sides AB and AC of triangle ABC, equilateral triangle ABD and ACE are drawn.

Prove that:

(i)Â âˆ CAD =Â âˆ BAEÂ  Â

(ii) CD = BE.

Solution:

Given triangle ABD is an equilateral triangle

Triangle ACE is an equilateral triangle

Now, we need to prove that

(i)Â âˆ CAD =Â âˆ BAEÂ  Â

(ii) CD = BE.

Proof:

(i) âˆ† ABD is equilateral

Each angle = 60o

Similarly,

âˆ† ACE is equilateral

Each angle = 60o

âˆ CAE =Â 60o â€¦â€¦â€¦..(ii)

âˆ BAD = âˆ CAE from (i) and (ii)â€¦â€¦â€¦..(iii)

Adding âˆ BAC to both sides, we have

âˆ BAD + âˆ BAC = âˆ CAE + âˆ BAC

(ii) In âˆ† CAD andÂ âˆ†Â BAE

AC = AE (triangle ACE is equilateral)

âˆ CAD = âˆ BAE from (iv)

AD = AB (triangle ABD is equilateral)

By SAS postulate of congruent triangles

The corresponding parts of the congruent triangles are congruent

Therefore CD = BE

Hence the proof.

2. In the following diagrams, ABCD is a square and APB is an equilateral triangle.

In each case,

(i) Prove that:Â âˆ† APDÂ â‰… âˆ† BPC

(ii) Find the angles ofÂ âˆ†DPC.

Solution:

(a)

(i) Proof:

AP = PB = AB [âˆ† APB is an equilateral triangle]

Also we have,

âˆ PBA = âˆ PAB = âˆ APB = 60o â€¦â€¦.. (1)

Since ABCD is a square, we have

âˆ A = âˆ B = âˆ C = âˆ D = 90o â€¦â€¦â€¦.. (2)

Since âˆ DAP = âˆ A – âˆ PAB â€¦â€¦â€¦â€¦.. (3)

âˆ DAP = 90o â€“ 60o

âˆ DAP = 30o [from equation 1 and equation 2] â€¦â€¦â€¦â€¦â€¦ (4)

Similarly âˆ CBP = âˆ B – âˆ PBA

âˆ CBP = 90o â€“ 60o

âˆ CBP = 30o [from equation 1 and equation 2] â€¦â€¦â€¦â€¦ (5)

âˆ DAP = âˆ CBP [from equation 4 and equation 5] â€¦â€¦. (6)

âˆ† APD and âˆ† BPC

AD = BC [sides of square ABCD]

âˆ DAP = âˆ CBP [from 6]

AP = BP [sides of equilateral triangle APB]

Therefore by SAS criteria of congruency, we have

âˆ† APDÂ â‰… âˆ† BPC

(ii) AP = PB = AB [âˆ† APB is an equilateral triangle] â€¦â€¦â€¦ (7)

AB = BC = CD = DA [sides of square ABCD] â€¦â€¦â€¦â€¦.. (8)

From equation 7 and 8, we have

AP = DA and PB = BC â€¦â€¦â€¦ (9)

In âˆ† APD,

AP = DA [from 9]

âˆ ADP = âˆ APD [angles opposite to equal sides are equal]

âˆ ADP + âˆ APD + âˆ DAP = 180o [sum of angles of a triangle = 180o]

âˆ ADP + âˆ APD + 30o = 180o

âˆ ADP + âˆ ADP = 180o â€“ 30o [from 2 and from 10]

We have âˆ PCD = âˆ C – âˆ PCB

âˆ PCD = 90o â€“ 75o

âˆ PCD = 15o â€¦â€¦â€¦ (13)

In triangle DPC

âˆ PDC = 15o

âˆ PCD = 15o

âˆ PCD + âˆ PDC + âˆ DPC = 180o

âˆ DPC = 180o â€“ 30o

âˆ DPC = 150o

Therefore angles are 15o, 150o and 15o

(b)

(i) Proof: In triangle APB

AP = PB = AB

Also,

We have,

âˆ PBA = âˆ PAB = âˆ APB = 60o â€¦â€¦â€¦ (1)

Since ABCD is a square, we have

âˆ A = âˆ B = âˆ C = âˆ D = 90o â€¦â€¦â€¦. (2)

âˆ DAP = âˆ A + âˆ PABâ€¦â€¦.. (3)

âˆ DAP = 90o + 60o

âˆ DAP = 150o [from 1 and 2] â€¦â€¦. (4)

âˆ CBP = âˆ B + âˆ PBAâ€¦â€¦.. (3)

âˆ CBP = 90o + 60o

âˆ CBP = 150o [from 1 and 2] â€¦â€¦ (5)

âˆ DAP = âˆ CBP [from 4 and 5] â€¦â€¦.. (6)

In triangle APD and triangle BPC

AD = BC [sides of square ABCD]

âˆ DAP = âˆ CBP [from 6]

AP = BP [sides of equilateral triangle APB]

By SAS criteria we have

âˆ† APDÂ â‰… âˆ† BPC

(ii) AP = PB = AB [triangle APB is an equilateral triangle] â€¦â€¦â€¦ (7)

AB = BC = CD = DA [sides of square ABCD] â€¦â€¦â€¦â€¦.. (8)

From equation 7 and 8, we have

AP = DA and PB = BC â€¦â€¦â€¦ (9)

In âˆ† APD,

AP = DA [from 9]

âˆ ADP = âˆ APD [angles opposite to equal sides are equal] â€¦â€¦â€¦.. (10)

âˆ ADP + âˆ APD + âˆ DAP = 180o [sum of angles of a triangle = 180o]

âˆ ADP + âˆ APD + 150o = 180o

âˆ ADP + âˆ ADP = 180o â€“ 150o [from 2 and from 10]

We have âˆ PCD = âˆ D – âˆ ADP

âˆ PCD = 90o â€“ 15o

âˆ PCD = 75o â€¦â€¦â€¦ (11)

In triangle BPC

PB = BC [from 9]

âˆ PCB = âˆ BPC â€¦â€¦â€¦ (12)

âˆ PCB + âˆ BPC + âˆ CPB = 180o

âˆ PCB + âˆ PCB = 180o â€“ 150o [from 2 and from 10]

2 âˆ PCB = 30o

âˆ PCB = 15o

We have âˆ PCD = âˆ C – âˆ PCB

âˆ PCD = 90o â€“ 15o

âˆ PCD = 75o â€¦â€¦â€¦ (11)

In triangle DPC

âˆ PDC = 75o

âˆ PCD = 75o

âˆ PCD + âˆ PDC + âˆ DPC = 180o

75o + 75o + âˆ DPC = 180o

âˆ DPC = 180o â€“ 150o

âˆ DPC = 30o

Angles of triangle are 75o, 30o and 75o

3. In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares. Prove that:

(i)Â âˆ†ACQ andÂ âˆ†ASB are congruent.

(ii) CQ = BS.

Solution:

Triangle ABC is right-angled at B.

ABPQ and ACRS are squares.

We need to prove that:

(i)Â âˆ†ACQ andÂ âˆ†ASB are congruent.

(ii) CQ = BS.

Proof:

(i) âˆ QAB = 90o (ABPQ is a square) â€¦. (1)

âˆ SAC = 90o (ACRS is a square) â€¦â€¦â€¦. (2)

From (1) and (2) we have

âˆ QAB = âˆ SAC â€¦â€¦â€¦â€¦ (3)

Adding âˆ BAC both sides of (3) we get

âˆ QAB + âˆ BAC = âˆ SAC + âˆ BAC

âˆ QAC = âˆ SAB â€¦â€¦â€¦. (4)

In âˆ† ACQ andÂ âˆ†Â ASB

QA = QB (sides of a square ABPQ)

âˆ CAD = âˆ BAE from (iv)

AC = AS (side of a square ACRS)

By AAS postulate of congruent triangles

Therefore âˆ† ACQ â‰… âˆ† ASB

(ii)The corresponding parts of the congruent triangles are congruent

Therefore CQ = BS

4. In aÂ âˆ†ABC, BD is the median to the side AC, BD is produced to E such that BD = DE. Prove that: AE is parallel to BC.

Solution:

Given in aÂ âˆ†ABC, BD is the median to the side AC,

BD is produced to E such that BD = DE.

Now we have to prove that: AE is parallel to BC.

Construction: Join AE

Proof:

AD = DC (BD is median to AC)

In âˆ† BDC andÂ âˆ†Â ADE

BD = DE (Given)

âˆ BDC = âˆ ADE = 90o (vertically opposite angles)

By SAS postulate of congruent triangles

Therefore âˆ† BDC â‰… âˆ† ADE

The corresponding parts of the congruent triangles are congruent

But these are alternate angles

And AC is the transversal

Thus, AE parallel to BC

5. In the adjoining figure, OX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.

If XSÂ âŠ¥ QR and XTÂ âŠ¥ PQ; prove that:

(i)Â âˆ† XTQ â‰… âˆ† XSQ

(ii) PX bisects angle âˆ P.

Solution:

OX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.

If XSÂ âŠ¥ QR and XTÂ âŠ¥ PQ;

We have to prove that:

(i)Â âˆ† XTQ â‰… âˆ† XSQ

(ii) PX bisects angle âˆ P.

Construction:

Draw If XZâŠ¥ PR and join PX

Proof:

(i) In âˆ† XTQ andÂ âˆ†Â XSQ

âˆ QTX = âˆ QSX = 90o (XS perpendicular to QR and XT perpendicular to PQ)

âˆ QTX = âˆ QSX (QX is bisector of angle Q)

QX = QX (common)

By AAS postulate of congruent triangles

Therefore âˆ† XTQ â‰… âˆ† XSQ â€¦â€¦â€¦â€¦.. (1)

(ii) The corresponding parts of the congruent triangles are congruent

Therefore XT = XS (by c.p.c.t)

In âˆ† XSR andÂ âˆ†Â XZR

âˆ XSR = âˆ XZR = 90o (XS perpendicular to XS and angle XSR = 90o)

âˆ SRX = âˆ ZRX (RX is a bisector of angle R)

RX = RX (common)

By AAS postulate of congruent triangles

Therefore âˆ† XSR â‰… âˆ† XZR â€¦â€¦â€¦â€¦.. (1)

The corresponding parts of the congruent triangles are congruent

Therefore XS = XZ (by c.p.c.t) â€¦â€¦â€¦.. (2)

From (1) and (2)

XT = XZ â€¦â€¦â€¦â€¦â€¦ (3)

In âˆ† XTP andÂ âˆ†Â XZP

âˆ XTP = âˆ XZP = 90o (Given)

XP = XP (common)

XT = XZ (from 3)

By right angle hypotenuse side postulate of congruent triangles

Therefore âˆ† XTP â‰… âˆ† XZP

The corresponding parts of the congruent triangles are congruent

âˆ XPT = âˆ XPZ

PX bisects âˆ SRX = âˆ P

6. In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.

Prove that: XA = YC.

Solution:

ABCD is a parallelogram in which âˆ A and âˆ C are obtuse.

Points X and Y are on the diagonal BD

Such that âˆ XAD = âˆ YCB = 90o

We need to prove that XA = YC

Proof:

In âˆ† XAD andÂ âˆ†Â YCB

âˆ XAD = âˆ YCB = 90o (Given)

AD = BC (opposite sides of a parallelogram)

âˆ ADX = âˆ CBY (alternate angles)

By ASA postulate of congruent triangles

Therefore âˆ† XAD â‰… âˆ† YCB

The corresponding parts of the congruent triangles are congruent

Therefore XA = YC

Hence the proof.

7. ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively, such produced to E and F respectively, such that AB = BE and AD = DF.

Prove that: âˆ† BEC â‰… âˆ† DCF

Solution:

ABCD is a parallelogram.

The sides AB and AD are produced to E and F respectively,

Such that AB = BE and AD = DF.

We need to prove that âˆ† BEC â‰… âˆ† DCF

Proof:

AB = DC (opposite sides of a parallelogram) â€¦â€¦. (1)

AB = BE (given) â€¦â€¦â€¦.. (2)

From (1) and (2) we have

BE = DC (opposite sides of a parallelogram) â€¦â€¦â€¦. (3)

AD = BC (opposite sides of a parallelogram) â€¦â€¦.. (4)

AD = DF (given) â€¦â€¦â€¦â€¦ (5)

From (4) and (5) we have

BC = DF â€¦â€¦â€¦ (6)

Since AD parallel to BC the corresponding angles are equal

âˆ DAB = âˆ CBE â€¦â€¦.. (7)

Since AD parallel to DC the corresponding angles are equal

âˆ DAB = âˆ FDC â€¦â€¦â€¦ (8)

From (7) and (8)

âˆ CBE = âˆ FDC â€¦â€¦.. (9)

In âˆ† BEC and âˆ† DCF

BE = DC (from (3))

âˆ CBE = âˆ FDC (from (9))

BC = DF (from (6))

By SAS postulate of congruent triangles

Therefore âˆ† BEC â‰… âˆ† DCF

Hence the proof.

8. In the following figures, the sides AB and BC and the median AD of triangle ABC are equal to the sides PQ and QR and median PS of the triangle PQR. Prove thatÂ âˆ†ABC andÂ âˆ†PQR are congruent.

Solution:

Since BC = QR

We have BD = QS and DC = SR (D is the midpoint of BC and S is the midpoint of QR)

In âˆ† ABD and âˆ† PQS

AB = PQ â€¦â€¦â€¦ (1)

BD = QS â€¦â€¦â€¦.. (3)

By SSS postulate of congruent triangles

Therefore âˆ† ABD â‰… âˆ† PQS

Similarly

In âˆ† ADC and âˆ† PSR

AC = PR â€¦â€¦â€¦.. (5)

DC = SR â€¦â€¦â€¦. (6)

By SSS postulate of congruent triangles

Therefore âˆ† ADC â‰… âˆ† PSR

We have

BC = BD + DC (D is the midpoint of BC)

= QS + SR (from (3) and (6))

= QR (S is the midpoint of QR) â€¦â€¦â€¦ (7)

Now again consider the triangles âˆ† ABC and âˆ† PR

AB = PQ (from 1)

BC = QR (from 7)

AC = PR (from 7)

By SSS postulate of congruent triangles

Therefore âˆ† ABC â‰… âˆ† PQR

Hence the proof.

9. In the following diagram, AP and BQ are equal and parallel to each other.

Prove that

(i) âˆ† AOP â‰… âˆ† BOQ

(ii) AB and PQ bisect each other

Solution:

In the figure AP and BQ are equal and parallel to each other

Therefore AP = BQ and AP parallel to BQ

We need to prove that

(i) âˆ† AOP â‰… âˆ† BOQ

(ii) AB and PQ bisect each other

(i) since AP parallel to BQ

âˆ APO = âˆ BQO (alternate angles) â€¦â€¦.. (1)

And âˆ PAO = âˆ QBO (alternate angles) â€¦â€¦.. (2)

Now in âˆ† AOP and âˆ† BOQ

âˆ APO = âˆ BQO (from 1)

AP = PQ (given)

âˆ PAO = âˆ QBO (from 2)

By ASA postulate of congruent triangles,

âˆ† AOP â‰… âˆ† BOQ

(ii) the corresponding parts of the congruent triangles are congruent

Therefore OP = OQ (by c.p.c.t)

OA = OB (by c.p.c.t)

Hence AB and PQ bisect each other

10. In the following figure, OA = OC and AB = BC.

(i) âˆ P= 90o

(ii) âˆ† AOD â‰… âˆ† COD

Solution:

Given OA = OC and AB = BC.

Now we have to prove that,

(i) âˆ P= 90o

(ii) âˆ† AOD â‰… âˆ† COD

(i) In âˆ† ABO and âˆ† CBO

AB = BC (given)

AO = CO (given)

OB = OB (common)

By SSS postulate of congruent triangles

Therefore âˆ† ABO â‰… âˆ† CBO

The corresponding parts of the congruent triangles are congruent

âˆ ABO = âˆ CBO (by c.p.c.t)Henceâˆ ABD = âˆ CBD

âˆ AOB = âˆ CBO (by c.p.c.t)

We have âˆ ABO + âˆ CBO = 180o (linear pair)

âˆ ABO = âˆ CBO = 90o

And AC perpendicular to BD

(ii) In âˆ† AOD and âˆ† COD

OD = OD (common)

âˆ AOD = âˆ COD (each = 90o)

AO = CO (given)

By SAS postulate of congruent triangles

Therefore âˆ† AOD â‰… âˆ† COD

(iii) The corresponding parts of the congruent triangles are congruent

Therefore AD = CD (by c.p.c.t)

Hence the proof.

## Selina Solutions for Class 9 Maths chapter 9- Triangles

The chapter 9, Triangles, contains 2 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

9.1 Introduction

9.2 Relation between sides and angles of triangles

9.3 Some important terms

9.4 Congruent Triangles

9.5 Conditions for congruency of triangles

## Selina Solutions for Class 9 Maths chapter 9- Triangles

The chapter 9 of class 9 deals about triangles. A triangle is a plane figure bounded by three straight line segments. Every triangle has three vertices and three sides.the chapter also explains about different terms like median, centroid, altitud, orthocentre etc. Read and learn the chapter 9 of Selina textbook to get to know more about Triangles. Learn the Selina Solutions for Class 9 effectively to attain excellent result in the examination.