# Triangles Concise Class-9th ICSE Maths Selina Publications

**Triangles Concise Class-9th ICSE** Mathematics Selina Publications Solutions Chapter-9 (Congruency in Triangles). We provide step by step Solutions of Exercise / lesson-9 **Triangles** for **ICSE** **Class-9 Concise** Selina Mathematics by R K Bansal.

Our Solutions contain all type Questions with Exe-9 A and Exe-9 B, to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-9 Mathematics .

**Triangles Concise Class-9th ICSE** Mathematics Selina Publications Solutions Chapter-9 (Congruency in Triangles).

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**Exercise – 9 A, Triangles Concise Class-9th ICSE Mathematics Selina Publications Solutions (Congruency in Triangles)**

#### Question 1

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

(a) In ΔABC and ΔDEF, AB = DE, BC = EF and B = E.

(b) In ΔABC and ΔDEF, B = E = 90^{o}; AC = DF and BC = EF.

(c) In ΔABC and Δ QRP, AB = QR, B = R and C = P.

(d) In ΔABC and Δ PQR, AB = PQ, AC = PR and BC = QR.

(e) In ΔADC and Δ PQRΔ, BC = QR, A = 90^{o}, C = R = 40^{o} and Q = 50^{o}.

( f) In Triangle ABC and Triangle BCD ………………………..

#### Answer

#### Question 2

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

…………….

Show that OP is perpendicular to AB

#### Answer

#### Question 3

The following figure shows a circle with centre O.

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

#### Answer

#### Question 4

In a triangle ABC, D is mid-point of BC; AD is produced upto E so that DE = AD. Prove that:

(i) ABD and ECD are congruent.

(ii) AB = CE.

(iii) AB is parallel to EC.

#### Answer

#### Question 5

A triangle ABC has B = 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.

#### Answer

#### Question 6

The perpendicular bisectors of the sides of a triangle ABC meet at I.

Prove that: IA = IB = IC.

#### Answer

#### Question 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.

#### Answer

#### Question 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.

#### Answer

#### Question 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) DCE LBE

(ii) AB = BL.

(iii) AL = 2DC

#### Answer

#### Question 10

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

If ABD = 58^{o},

DBC = (2x – 4)^{o},

ACB = y + 15^{o} and

DCB = 63^{o} ; find the values of x and y.

#### Answer

#### Question 11

In the given figure: AB//FD, AC//GE and BD = CE; prove that:

(i) BG = DF

(ii) CF = EG

#### Answer

#### Question 12

In ∆ABC, AB = AC. Show that the altitude AD is median also.

#### Answer

#### Question 13

In the following figure, BL = CM.

Prove that AD is a median of triangle ABC.

#### Answer

#### Question 14

In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.

Prove that :

(i) BD = CD

(ii) ED = EF

#### Answer

#### Question 15

Use the information in the given figure to prove :

(i) AB = FE

(ii) BD = CF

#### Answer

### Selina Publications Solutions **Triangles Concise Class-9th ICSE** Mathematics (Congruency in Triangles) **Exercise – 9 B**

#### Question 1

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

Prove that: (i) CAD = BAE (ii) CD = BE.

#### Answer

#### Question 2

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

In each case,

(i) Prove that: .ΔAPD = Δ BPC

(ii) Find the angles of .Δ DPC

#### Answer

#### Question 3

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

(i) .Δ ACQ and .Δ ASB are congruent.

(ii) CQ = BS

#### Answer

#### Question 4

In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE. Prove that: AE is parallel to BC.

#### Answer

#### Question 5

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

If XS QR and XT PQ ; prove that:

(i) ……………………..

(ii) PX bisects angle P.

#### Answer

#### Question 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.

#### Answer

#### Question 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: Δ BEC=Δ DCF.

#### Answer

Hence proved

#### Question 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 that ABC and PQR are congruent.

#### Answer

#### Question 9

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

Prove that:

(i) ……………………

(ii) AB and PQ bisect each other.

#### Answer

#### Question 10

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

Prove that:

(i) AOB = 90^{o}

(ii) triangle AOD =triangle COD

(iii) AD = CD

#### Answer

#### Question 11

The following figure shown a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN.

Prove that:

(i) AM = AN (ii)triangle AMC = triangle ANB

(iii) BN = CM (iv) triangle BMC = triangle CNB

#### Answer

#### Question 12

In a triangle ABC, AB = BC, AD is perpendicular to side BC and CE is perpendicular to side AB. Prove that : AD = CE.

#### Answer

#### Question 13

PQRS is a parallelogram. L and M are points on PQ and SR respectively such that PL= MR. Show that LM and QS bisect each other.

#### Answer

#### Question 14

In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced upto point R so that CR = BP.

Prove that QR bisects PC.

Hint: (Show that ∆ QBP is equilateral

⇒ BP = PQ, but BP = CR

⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM)

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