Chapter 9 : Triangles [Congruency in Triangles] - Selina Solutions for Class 9 Maths ICSE
Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.
Chapter 9 - Triangles [Congruency in Triangles] Excercise Ex. 9(A)
Which of the following pairs of triangles are congruent? In each case, state the condition of congruency:
(a) In ABC and DEF, AB = DE, BC = EF and B = E.
(b) In ABC and DEF, B = E = 90o; AC = DF and BC = EF.
(c) In ABC and QRP, AB = QR, B = R and C = P.
(d) In ABC and PQR, AB = PQ, AC = PR and BC = QR.
(e) In ADC and PQR, BC = QR, A = 90o, C = R = 40o and Q = 50o.
The given figure shows a circle with centre O. P is mid-point of chord AB.
Show that OP is perpendicular to AB.
The following figure shows a circle with centre O.
If OP is perpendicular to AB, prove that AP = BP.
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.
A triangle ABC has B = C.
(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 bisectors of the sides of a triangle ABC 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 al line segment and E is mid point of BC.
(i) DCE LBE
(ii) AB = BL.
(iii) AL = 2DC
In the given figure, AB = DB and Ac = DC.
If ABD = 58o,
DBC = (2x - 4)o,
ACB = y + 15o and
DCB = 63o ; find the values of x and y.
In the given figure: AB//FD, AC//GE and BD = CE; prove that:
(i) BG = DF
(ii) CF = EG
In ∆ABC, AB = AC. Show that the altitude AD is median also.
In the following figure, BL = CM.
Prove that AD is a median of triangle ABC.
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
Chapter 9 - Triangles [Congruency in Triangles] Excercise Ex. 9(B)
On the sides AB and AC of triangle ABC, equilateral triangle ABD and ACE are drawn.
Prove that: (i) CAD = BAE (ii) CD = BE.
In the following diagrams, ABCD is a square and APB is an equilateral triangle.
In each case,
(i) Prove that: APD BPC
(ii) Find the angles of DPC.
In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares. Prove that:
(i) ACQ and ASB are congruent.
(ii) CQ = BS.
In a 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, 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:
(ii) PX bisects angle P.
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.
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: BEC DCF.
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.
In the following diagram, AP and BQ are equal and parallel to each other.
(ii) AB and PQ bisect each other.
In the following figure, OA = OC and AB = BC.
(i) AOB = 90o
(ii) AOD COD
(iii) AD = CD
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.
(i) AM = AN (ii) AMCANB
(iii) BN = CM (iv) BMC CNB
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 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)
In the following figure, ∠A = ∠C and AB = BC. Prove that ΔABD ≅ ΔCBE.
In triangles AOE and COD,
∠A = ∠C (given)
∠AOE = ∠COD (vertically opposite angles)
∴ ∠A + ∠AOE = ∠C + ∠COD
⇒180° - ∠AEO = 180° - ∠CDO
⇒ ∠AEO = ∠ CDO ….(i)
Now, ∠AEO + ∠OEB = 180° (linear pair)
And, ∠CDO + ∠ODB = 180° (linear pair)
∴ ∠AEO + ∠OEB = ∠CDO + ∠ODB
⇒ ∠OEB = ∠ODB [Using (i)]
⇒ ∠CEB = ∠ADB ….(ii)
Now, in ΔABD and ΔCBE,
∠A = ∠C (given)
∠ADB = ∠CEB [From (ii)]
AB = BC (given)
⇒ ΔABD ≅ ΔCBE (by AAS congruence criterion)
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 ΔAOD and ΔBOC,
∠ AOD = ∠ BOC (vertically opposite angles)
∠ DAO = ∠ CBO (each 90°)
AD = BC (given)
∴ ΔAOD ≅ ΔBOC (by AAS congruence criterion)
⇒ AO = BO [cpct]
⇒ O is the mid-point of AB.
Hence, 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.
AB = AC
⇒ ∠B = ∠C (angles opposite to equal sides are equal)
In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.
Prove that AD = FC.
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
In ΔABD and ΔBAC,
AD = BC (given)
BD = CA (given)
AB = AB (common)
∴ ΔABD ≅ ΔBAC (by SSS congruence criterion)
⇒ ∠ADB = ∠BCA and ∠DAB = ∠CBA (cpct)
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