Selina solutions concise maths class 9 icse Solutions for Chapter 9 chapter 9 triangles congruency in triangles

Exercises in Chapter 9 chapter 9 triangles congruency in triangles Grade 9

Exercise 9(A)

Exercise 9(B)

Questions in Exercise 9(A)

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

condition of congruency:

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

(b) In 𝚫 ABC and 𝚫 DEF, ∠B=∠E = 90° ; AC=DF and BC=EF.

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

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

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

The following figure shows a circle with center O.

In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE =

AD. Prove that:

(i) 𝚫 ABD and 𝚫 ECD are congruent.

(ii) AB=EC

(iii) AB is parallel to EC.

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.

The perpendicular bisector of the sides of a triangle AB meet at I.

Prove that: IA=IB=IC.

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.

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.

From the given diagram, in which ABCD is a parallelogram, ABL is a line

segment and E is mid-point of BC.

Prove that:

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

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

(i) BG=DF

(ii) CF=EG

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

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

Use the information in the given figure to prove:

(i) AB=FE

(ii) BD=CF

Questions in Exercise 9(B)

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

drawn. Prove that:

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

In each case,

In the figure, given below, triangle ABC is a right-angled triangle at B. ABPQ and

ACRS are squares. Prove that:

(i) 𝚫ACQ and 𝚫ASB are congruent.

(ii) CQ=BS

In a 𝚫 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.

In the adjoining figure, QX and RX are the bisectors of the angles Q and R

respectively of the triangle PQR.

If XS⊥QR and XT⊥PQ; prove that:

In 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.

ABCD is a parallelogram. The sides AB and AD are produced to E and F

respectively, such that AB=BE and AD=DF.

Prove that: 𝚫BEC≅ 𝚫DCF

In the following figures, the sides AB and BC and the median AD of triangle

ABC are respectively equal to the sides PQ and QR and the median PS of the

triangle PQR. Prove that 𝚫ABC and 𝚫PQR are congruent.

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

Prove that:

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

The following figure shows 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) 𝚫AMC ≅ 𝚫ANB

(iii) BN=CM

(iv) 𝚫BMC ≅ 𝚫CNB

In a triangle, ABC, AB=BC, AD is perpendicular to side BC and CE is

perpendicular to side AB. Prove that:

AD=CE.

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.

In the following figure, ABC is an equilateral triangle in which P is parallel to

AC. Side AC is produced up to point R so that CR=BP. Prove that QR bisects

PC.

In the following figure ∠A=∠C and AB=BC.

AD and BC are equal perpendiculars to a line segment AB. If AD and BC are

on different sides of AB prove that CD bisects AB.

In 𝚫ABC, AB=AC and the bisectors of angles B and C intersect at point O.

Prove that:

(i) BO=CO

(ii) AO bisects angle BAC.

A point O is taken inside a rhombus ABCD such that its distance from the

vertices B and D are equal. Show that AOC is a straight line.

In quadrilateral ABCD, AD=BC and BD=CA. Prove that:

(i) ∠ADB = ∠BCA

(ii) ∠DAB = ∠CBA

Was This helpful?

Chapters in this book

Chapter 1- Rational and Irrational Numbers

Chapter 2- Compound Interest [Without Using Formula

Chapter 3- Compound Interest [Using Formula

Chapter 4- Expansions

Chapter 5- Factorisation

Chapter 6- Simultaneous Equations

Chapter 7- Indices [exponents]

Chapter 8- Logarithms

Chapter 9- Triangles [Congruency in Triangles]

Chapter 10- Isosceles Triangle

Chapter 11- Inequalities

Chapter 12- Mid-Point and Its Converse

Chapter 13- Pythagoras Theorem

Chapter 14- Rectilinear Figures

Chapter 15- Construction of Polygons

Chapter 16- Area Theorems

Chapter 17- Circles

Chapter 18- Statistics

Chapter 19- Mean and Median

Chapter 20- Area and Perimeter of Plane Figures

Chapter 21- Solids

Chapter 22- Trigonometrical Ratios

Chapter 23- Trigonometrical Ratios of Standard Angles

Chapter 24- Solution Of Right Triangles

Chapter 25- Complementary angles

Chapter 26- Co-ordinate Geometry

Chapter 27- Graphical Solution

Chapter 28- Distance Formula