TRIANGLES [Congruency in Triangles] - Questions & Answers

EXERCISE 9(A)

1. Multiple Choice Type :

Choose the correct answer from the options given below.
(a) If ΔABC ≅ ΔPQR, then
(i) AC = PR
(ii) AC = PQ
(iii) BC = PR
(iv) BC = PQ
Answer:
(i) AC = PR
(By Corresponding Parts of Congruent Triangles, abbreviated as CPCTC, the corresponding sides are equal.)

(b) Which of the following will hold true for the given figure :
(i) AD = DC
(ii) CD = CB
(iii) ∠ACD ≠ ∠ACB
(iv) ∠B ≠ ∠D
Answer:
(ii) CD = CB
(Based on the geometrical markings provided in the figure, sides with double tick marks are equal.)

(c) In the given figure, AM is the perpendicular bisector of BC. Then :
(i) AB = AM
(ii) AC = BM
(iii) AB ≠ AC
(iv) AM bisects ∠BAC
Answer:
(iv) AM bisects ∠BAC
In ΔAMB and ΔAMC:
BM = MC (Since AM is the bisector of BC)
∠AMB = ∠AMC = 90° (Since AM is perpendicular to BC)
AM = AM (Common side)
∴ ΔAMB ≅ ΔAMC (By SAS congruency)
∴ ∠BAM = ∠CAM (By CPCTC)
Hence, AM bisects ∠BAC.

(d) Which of the following is true for the given figure :
(i) ΔAPC ≅ ΔBPD
(ii) CP = DP
(iii) AB and CD bisect each other
(iv) all of the above are true
Answer:
(iv) all of the above are true
In ΔAPC and ΔBPD:
AP = BP (Given by double tick marks)
∠PAC = ∠PBD = 90° (Given)
AC = BD (Given by single tick marks)
∴ ΔAPC ≅ ΔBPD (By SAS congruency, option i is true)
∴ CP = DP (By CPCTC, option ii is true)
Since AP = BP and CP = DP, the lines bisect each other at P (option iii is true).

(e) In the following figure, ∠BAD = ∠EAC, BD = EC and ∠B = ∠E, then :
(i) ΔABD ¬≅ ΔAEC
(ii) ΔABC ≅ ΔAED
(iii) ΔABC ¬≅ ΔAED
(iv) ΔABD ≅ ΔADE
Answer:
(ii) ΔABC ≅ ΔAED
Given ∠BAD = ∠EAC.
Adding ∠DAC to both sides: ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC = ∠EAD.
Applying AAS and CPCTC conditions logically leads to the congruency of the larger triangles.

(f) Which of the following is true for the given figure :
(i) ΔABD ≅ ΔACD
(ii) angle BAD ≠ angle CAD
(iii) ΔABD ¬≅ ΔACD
(iv) ∠EAB = ∠BAD
Answer:
(i) ΔABD ≅ ΔACD
(Based on corresponding symmetrical parts).

(g) In the given figure, ∠x = ∠y and PO = RO, then :
(i) RB = AO
(ii) BO = PA
(iii) BP = AR
(iv) RB = OB
Answer:
(i) RB = AO
In ΔPAO and ΔRBO:
∠A = ∠B = 90° (Given)
∠x = ∠y (Given)
PO = RO (Given)
∴ ΔPAO ≅ ΔRBO (By AAS congruency)
∴ AO = RB (By CPCTC)

(h) In the given figure, BC // DA and BC = DA, then :
(i) AB and CD bisect each other
(ii) AB ≠ CD
(iii) OA = OC
(iv) OA = OD
Answer:
(i) AB and CD bisect each other
In ΔOBC and ΔODA:
∠BOC = ∠DOA (Vertically opposite angles)
∠OBC = ∠ODA (Alternate interior angles since BC // DA)
BC = DA (Given)
∴ ΔOBC ≅ ΔODA (By AAS congruency)
∴ OB = OD and OC = OA (By CPCTC)
Thus, the diagonals AB and CD bisect each other at O.

2. The following figure shows a circle with centre O. If OP is perpendicular to AB, prove that AP = BP.
Answer:
Given: A circle with center O, and OP is perpendicular to chord AB.
To Prove: AP = BP
Construction: Join OA and OB.
In right-angled ΔOPA and ΔOPB:
OA = OB (Radii of the same circle)
OP = OP (Common side)
∠OPA = ∠OPB = 90° (Given, OP is perpendicular to AB)
∴ ΔOPA ≅ ΔOPB (By RHS congruency criterion)
∴ AP = BP (By CPCTC)
Hence proved.

3. In a triangle ABC, D is mid-point of BC; AD is produced upto E so that DE = AD. Prove that :
(i) ΔABD and ΔECD are congruent.
(ii) AB = EC.
(iii) AB is parallel to EC.
Answer:
(i) In ΔABD and ΔECD:
BD = CD (Given, D is the mid-point of BC)
∠ADB = ∠CDE (Vertically opposite angles)
AD = ED (Given)
∴ ΔABD ≅ ΔECD (By SAS congruency criterion)
Hence proved.

(ii) Since ΔABD ≅ ΔECD (Proved in part i):
AB = EC (By CPCTC)
Hence proved.

(iii) Since ΔABD ≅ ΔECD:
∠DAB = ∠DEC (By CPCTC)
These are alternate interior angles formed by transversal AE intersecting lines AB and EC.
Since alternate interior angles are equal, AB is parallel to EC.
Hence proved.

4. From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid point of BC. Prove that :
(i) ΔDCE ≅ ΔLBE
(ii) AB = BL.
(iii) AL = 2DC
Answer:
(i) In ΔDCE and ΔLBE:
∠DEC = ∠LEB (Vertically opposite angles)
CE = BE (Given, E is the mid-point of BC)
∠DCE = ∠LBE (Alternate interior angles, as DC // AL and BC is a transversal)
∴ ΔDCE ≅ ΔLBE (By ASA congruency criterion)
Hence proved.

(ii) Since ΔDCE ≅ ΔLBE (Proved in part i):
DC = BL (By CPCTC)
Also, ABCD is a parallelogram, so opposite sides are equal: AB = DC.
Comparing the two equations: AB = BL.
Hence proved.

(iii) From the figure, AL is a line segment consisting of AB and BL.
So, AL = AB + BL
Since AB = DC and BL = DC (Proved above), we can substitute:
AL = DC + DC
AL = 2DC
Hence proved.

5. On the sides AB and AC of triangle ABC, equilateral triangles ABD and ACE are drawn. Prove that: (i) ∠CAD = ∠BAE (ii) CD = BE.
Answer:
(i) Since ΔABD and ΔACE are equilateral triangles:
∠DAB = 60° and ∠EAC = 60° (Angles of an equilateral triangle)
Now, ∠CAD = ∠CAB + ∠DAB = ∠CAB + 60°
And, ∠BAE = ∠CAB + ∠EAC = ∠CAB + 60°
Therefore, ∠CAD = ∠BAE.
Hence proved.

(ii) In ΔADC and ΔABE:
AD = AB (Sides of the same equilateral ΔABD)
∠CAD = ∠BAE (Proved in part i)
AC = AE (Sides of the same equilateral ΔACE)
∴ ΔADC ≅ ΔABE (By SAS congruency criterion)
∴ CD = BE (By CPCTC)
Hence proved.

6. 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. (HOTS)
Answer:
Given: Line segment AB is bisected at P, which means AP = BP. PQ is perpendicular to AB, meaning ∠QPA = ∠QPB = 90°.
To show: QA = QB
In ΔQPA and ΔQPB:
AP = BP (Given, P is the mid-point of AB)
∠QPA = ∠QPB = 90° (Given, PQ ⊥ AB)
QP = QP (Common side)
∴ ΔQPA ≅ ΔQPB (By SAS congruency criterion)
∴ QA = QB (By CPCTC)
Hence proved.

7. In the following diagrams, ABCD is a square and APB is an equilateral triangle. In each case,
(i) Prove that : ΔAPD ≅ ΔBPC
(ii) Find the angles of ΔDPC.
Answer:
(i) In ΔAPD and ΔBPC:
AD = BC (Sides of square ABCD are equal)
AP = BP (Sides of equilateral triangle APB are equal)
The angle ∠DAP = ∠DAB + ∠PAB = 90° + 60° = 150°
The angle ∠CBP = ∠CBA + ∠PBA = 90° + 60° = 150°
∴ ∠DAP = ∠CBP
∴ ΔAPD ≅ ΔBPC (By SAS congruency criterion)
Hence proved.

(ii) Since ΔAPD ≅ ΔBPC, we get DP = CP (By CPCTC).
Thus, ΔDPC is an isosceles triangle.
In ΔAPD, AD = AB (Sides of square) and AP = AB (Sides of equilateral Δ).
So, AD = AP, making ΔAPD an isosceles triangle.
∠ADP = ∠APD = (180° - 150°) / 2 = 30° / 2 = 15°.
Similarly, in ΔBPC, ∠BCP = 15°.
Now, for the angles of ΔDPC:
∠PDC = ∠ADC - ∠ADP = 90° - 15° = 75°.
∠PCD = ∠BCD - ∠BCP = 90° - 15° = 75°.
∠DPC = 180° - (∠PDC + ∠PCD) = 180° - (75° + 75°) = 180° - 150° = 30°.
Therefore, the angles of ΔDPC are 75°, 75°, and 30°.

8. In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares. Prove that :
(i) ΔACQ and ΔASB are congruent.
(ii) CQ = BS.
Answer:
(i) In ΔACQ and ΔASB:
AC = AS (Sides of the same square ACRS)
AQ = AB (Sides of the same square ABPQ)
∠CAQ = ∠CAB + ∠QAB = ∠CAB + 90°
∠SAB = ∠CAB + ∠SAC = ∠CAB + 90°
∴ ∠CAQ = ∠SAB
∴ ΔACQ ≅ ΔASB (By SAS congruency criterion)
Hence proved.

(ii) Since ΔACQ ≅ ΔASB (Proved above):
CQ = BS (By CPCTC)
Hence proved.

9. 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:
In ΔADE and ΔCDB:
AD = CD (Given, BD is the median, so D is the mid-point of AC)
∠ADE = ∠CDB (Vertically opposite angles)
DE = BD (Given)
∴ ΔADE ≅ ΔCDB (By SAS congruency criterion)
∴ ∠DAE = ∠DCB (By CPCTC)
These are alternate interior angles formed by the transversal AC cutting lines AE and BC.
Since the alternate interior angles are equal, AE is parallel to BC.
Hence proved.

10. 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.
Answer:
Since ABCD is a parallelogram, opposite sides are equal: AB = CD and AD = BC.
Also, opposite angles are equal: ∠ABC = ∠ADC.
Given AB = BE, therefore CD = BE (Since AB = CD).
Given AD = DF, therefore BC = DF (Since AD = BC).
Now, ∠CBE = 180° - ∠ABC (Linear pair)
And ∠CDF = 180° - ∠ADC (Linear pair)
Since ∠ABC = ∠ADC, it follows that ∠CBE = ∠CDF.
In ΔBEC and ΔDCF:
BE = CD (Proved above)
∠CBE = ∠CDF (Proved above)
BC = DF (Proved above)
∴ ΔBEC ≅ ΔDCF (By SAS congruency criterion)
Hence proved.

11. 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.
Answer:
Given: ABC is an equilateral triangle. QP // AC, and CR = BP.
First, show that ΔQBP is an equilateral triangle:
Since QP // AC, ∠BQP = ∠BAC = 60° (Corresponding angles).
Also, ∠BPQ = ∠BCA = 60° (Corresponding angles).
In ΔQBP, all angles are 60°, so it is an equilateral triangle.
⇒ BP = PQ = QB.
Since we are given BP = CR, we can substitute to get PQ = CR.
Let the intersection of QR and PC be M.
In ΔQPM and ΔRCM:
∠QMP = ∠RMC (Vertically opposite angles)
∠PQM = ∠CRM (Alternate interior angles, since QP // AR, so QP // CR)
PQ = CR (Proved above)
∴ ΔQPM ≅ ΔRCM (By AAS congruency criterion)
∴ PM = CM (By CPCTC)
Since PM = CM, M is the midpoint of PC, meaning QR bisects PC.
Hence proved.

12. 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:
Let LM and QS intersect at point O.
Since PQRS is a parallelogram, PQ // SR.
Because L is on PQ and M is on SR, PL // MR.
In ΔPLO and ΔMRO:
∠POL = ∠MOR (Vertically opposite angles)
∠PLO = ∠RMO (Alternate interior angles, since PL // MR and LM is a transversal)
PL = MR (Given)
∴ ΔPLO ≅ ΔMRO (By AAS congruency criterion)
∴ LO = MO and PO = RO (By CPCTC)
Wait, the diagonals are QS and LM. So the triangles to consider are ΔLSO and ΔMQO, or similar.
Let's consider ΔLQO and ΔMSO:
PQ = SR (Opposite sides of parallelogram).
LQ = PQ - PL and SM = SR - MR.
Since PQ = SR and PL = MR, it implies LQ = SM.
In ΔLQO and ΔMSO:
∠LOQ = ∠MOS (Vertically opposite angles)
∠LQO = ∠MSO (Alternate interior angles, since PQ // SR)
LQ = SM (Proved above)
∴ ΔLQO ≅ ΔMSO (By AAS congruency criterion)
∴ LO = MO and QO = SO (By CPCTC).
This means O is the mid-point of both LM and QS.
Thus, LM and QS bisect each other.
Hence proved.

EXERCISE 9(B)

1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) In the given figure, AB = PQ, BC = QR and median AM = median PN, then :
(i) AC ≠ PR
(ii) BM ≠ QN
(iii) ΔABM ¬≅ ΔPQN
(iv) ΔABC ≅ ΔPQR
Answer:
(iv) ΔABC ≅ ΔPQR
(Since BC = QR, their halves BM = QN. Then ΔABM ≅ ΔPQN by SSS. This makes ∠B = ∠Q. Finally, ΔABC ≅ ΔPQR by SAS).

(b) In triangles ABC and DEF, AB = DE and AC = EF, then to makes these two triangles congruent, we must have :
(i) BC = DF
(ii) ∠A = ∠E
(iii) any of (i) and (ii)
(iv) none of (i) and (ii)
Answer:
(iii) any of (i) and (ii)
(If BC=DF, they are congruent by SSS. If ∠A=∠E, which are the included angles between the given sides, they are congruent by SAS).

(c) In quadrilateral ABCD, AB = AC and BD = CD, then AD bisects :
(i) angle ADC
(ii) angle BAD
(iii) angle BAC
(iv) angle ABC
Answer:
(iii) angle BAC
(Wait, by looking at the figure, AB = AC and BD = CD. In ΔABD and ΔACD, AD is common. They are congruent by SSS. Therefore, ∠BAD = ∠CAD, which means AD bisects ∠BAC. Option (iii) matches the full angle name BAC).

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:
Given: A circle with center O, and P is the mid-point of chord AB (so AP = BP).
To show: OP ⊥ AB.
Construction: Join OA and OB.
In ΔOAP and ΔOBP:
OA = OB (Radii of the same circle)
AP = BP (Given, P is the mid-point)
OP = OP (Common side)
∴ ΔOAP ≅ ΔOBP (By SSS congruency criterion)
∴ ∠OPA = ∠OPB (By CPCTC)
Since AB is a straight line, ∠OPA + ∠OPB = 180° (Linear pair).
∠OPA + ∠OPA = 180° ⇒ 2∠OPA = 180° ⇒ ∠OPA = 90°.
Therefore, OP is perpendicular to AB.
Hence proved.

3. 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 from B and C to the opposite sides are equal.
Answer:
Given: ΔABC where ∠B = ∠C.
(i) Let M be the mid-point of BC. Draw MD ⊥ AB and ME ⊥ AC.
In ΔMDB and ΔMEC:
∠MDB = ∠MEC = 90° (By construction)
∠B = ∠C (Given)
MB = MC (Since M is the mid-point of BC)
∴ ΔMDB ≅ ΔMEC (By AAS congruency criterion)
∴ MD = ME (By CPCTC).
Hence proved.

(ii) Draw perpendiculars BF ⊥ AC and CG ⊥ AB.
In ΔBFC and ΔCGB:
∠BFC = ∠CGB = 90° (By construction)
∠C = ∠B (Given)
BC = CB (Common side)
∴ ΔBFC ≅ ΔCGB (By AAS congruency criterion)
∴ BF = CG (By CPCTC).
Hence proved.

4. In the given 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 :
(i) ΔXTQ ≅ ΔXSQ
(ii) PX bisects angle P.
Answer:
(i) In ΔXTQ and ΔXSQ:
∠XTQ = ∠XSQ = 90° (Given XS ⊥ QR and XT ⊥ PQ)
∠TQX = ∠SQX (Given QX is the bisector of ∠Q)
QX = QX (Common side/Hypotenuse)
∴ ΔXTQ ≅ ΔXSQ (By AAS congruency criterion)
Hence proved.

(ii) From the congruency above, XT = XS (By CPCTC).
Draw XU ⊥ PR. Similarly, by taking ΔXSR and ΔXUR, we can prove ΔXSR ≅ ΔXUR (using the bisector of R).
This gives XS = XU (By CPCTC).
Since XT = XS and XS = XU, we get XT = XU.
Now, in ΔXTP and ΔXUP:
∠XTP = ∠XUP = 90°
XP = XP (Common hypotenuse)
XT = XU (Proved above)
∴ ΔXTP ≅ ΔXUP (By RHS congruency criterion)
∴ ∠TPX = ∠UPX (By CPCTC).
This means PX bisects angle P.
Hence proved.

5. 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 median PS of the triangle PQR. Prove that ΔABC and ΔPQR are congruent.
Answer:
Given: AB = PQ, BC = QR, and AD = PS. AD is median of BC (so BD = DC = BC/2) and PS is median of QR (so QS = SR = QR/2).
Since BC = QR, it follows that BC/2 = QR/2 ⇒ BD = QS.
In ΔABD and ΔPQS:
AB = PQ (Given)
BD = QS (Proved above)
AD = PS (Given)
∴ ΔABD ≅ ΔPQS (By SSS congruency criterion)
∴ ∠B = ∠Q (By CPCTC)
Now, in the large triangles ΔABC and ΔPQR:
AB = PQ (Given)
∠B = ∠Q (Proved above)
BC = QR (Given)
∴ ΔABC ≅ ΔPQR (By SAS congruency criterion)
Hence proved.

6. In the following figure, OA = OC and AB = BC. Prove that :
(i) ∠AOB = 90°
(ii) ΔAOD ≅ ΔCOD
(iii) AD = CD
Answer:
(i) In ΔAOB and ΔCOB:
OA = OC (Given)
AB = CB (Given)
OB = OB (Common side)
∴ ΔAOB ≅ ΔCOB (By SSS congruency criterion)
∴ ∠AOB = ∠COB (By CPCTC)
Since AC is a straight line, ∠AOB + ∠COB = 180°.
⇒ 2∠AOB = 180° ⇒ ∠AOB = 90°.
Hence proved.

(ii) Since ∠AOB = 90°, ∠AOD = 90° and ∠COD = 90° (Linear pairs on straight lines).
In ΔAOD and ΔCOD:
OA = OC (Given)
∠AOD = ∠COD = 90° (Proved above)
OD = OD (Common side)
∴ ΔAOD ≅ ΔCOD (By SAS congruency criterion).
Hence proved.

(iii) Since ΔAOD ≅ ΔCOD (Proved in part ii):
AD = CD (By CPCTC).
Hence proved.

7. 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
Answer:
(i) Given AB = AC and BM = CN.
AM = AB - BM
AN = AC - CN
Since AB = AC and BM = CN, substituting gives AM = AN.
Hence proved.

(ii) In ΔAMC and ΔANB:
AM = AN (Proved in part i)
∠A = ∠A (Common angle)
AC = AB (Given)
∴ ΔAMC ≅ ΔANB (By SAS congruency criterion)
Hence proved.

(iii) Since ΔAMC ≅ ΔANB (Proved in part ii):
CM = BN (By CPCTC).
Hence proved.

(iv) In ΔBMC and ΔCNB:
BM = CN (Given)
CM = BN (Proved in part iii)
BC = CB (Common base)
∴ ΔBMC ≅ ΔCNB (By SSS congruency criterion)
Hence proved.

8. In a triangle ABC, AB = BC, AD is perpendicular to side BC and CE is perpendicular to side AB. Prove that : AD = CE.
Answer:
In ΔABD and ΔCBE:
∠ADB = ∠CEB = 90° (Given, AD ⊥ BC and CE ⊥ AB)
∠B = ∠B (Common angle)
AB = CB (Given)
∴ ΔABD ≅ ΔCBE (By AAS congruency criterion)
∴ AD = CE (By CPCTC)
Hence proved.

TEST YOURSELF

1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. In order to make these triangles congruent, we must have AB equal to :
(i) PQ
(ii) PR
(iii) QR
(iv) none of these
Answer:
(iii) QR
(Because side AB lies between ∠A and ∠B, its corresponding side must lie between corresponding angles ∠Q and ∠R, which is QR).

(b) If two sides and an angle of one triangle are equal to two sides and an angle of the other triangle, then the triangles must be congruent.
(i) no
(ii) yes
(iii) can't say
Answer:
(i) no
(The angle MUST be the "included angle" between the two sides for the SAS criterion to hold true. SSA is not a valid congruency condition).

(c) If BA = DE, AC = DF and BF = EC, then the triangles ABC and DEF are congruent by axiom.
(i) ASA
(ii) AAS
(iii) RHS
(iv) SSS
Answer:
(iv) SSS
(Given BF = EC, add FC to both sides: BF + FC = EC + FC ⇒ BC = EF. Now we have BA=DE, AC=DF, and BC=EF, which is Side-Side-Side).

(d) If BM = DM then AM = CM :
(i) yes
(ii) no
(iii) can't say
(iv) none of these
Answer:
(iii) can't say
(Just knowing one pair of line segments is bisected by a point M does not guarantee the other intersecting line segment is also bisected without additional information like parallel lines or equal angles).

(e) Statement (1) : ∠A = ∠Q and ∠B = ∠R, then to get the triangles, congruent, we must have AB = PR.
Statement (2) : The given Δs will be congruent, if AB = QR.
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Answer:
(iv) Statement 1 is false, and statement 2 is true.
(As established in part a, AB must correspond to QR for ASA congruency).

(f) Statement (1) : MM′ is a plane mirror and A is an object, then I the image of object A in mirror MM′ and so IO = AO.
Statement (2) : ΔAOC ≅ ΔIOC by ASA. And, so IO = AO.
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Answer:
(i) Both the statements are true.
(In plane mirror reflection, image distance equals object distance. The geometric proof for this uses ASA congruency on the normal and incident/reflected rays).

(g) Statement (1) : If two angles and a side of one triangle are equal to two angles and a side of some another triangle, the triangles are congruent.
Statement (2) : The two triangles will be congruent, if corresponding sides of two triangles are equal.
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Answer:
(i) Both the statements are true.
(Statement 1 refers to AAS or ASA axioms. Statement 2 refers to the SSS axiom).

(h) Assertion (A) : If PQ = PR, ΔPQS ≅ ΔPRT.
Reason (R) : PQ = PR, ∠P = ∠P and ∠Q = ∠R
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Answer:
(ii) A is false, R is true.
(Wait, if PQ = PR in ΔPQR, then opposite angles are equal, so ∠PQR = ∠PRQ, which means ∠Q = ∠R is true. However, to prove ΔPQS ≅ ΔPRT we need more than just one side and two angles unless they strictly correspond. From the figure, T and S are just points on the lines. Without knowing QT=RS or similar, we can't definitively assert the triangles are congruent).

(i) Assertion (A) : ΔABD ≅ ΔACE.
Reason (R) : ∠ADE + ∠ADB = ∠AEC + ∠AED
But AD = AE ⇒ ∠ADE = ∠AED ∴ ∠ADB = ∠AEC ⇒ ΔABD ≅ ΔACE
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Answer:
(iii) Both A and R are true and R is the correct reason for A.
(The reason perfectly provides the step-by-step logic: supplementary angles to equal base angles of an isosceles triangle are also equal, leading to SAS congruency).

2. 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°; AC = DF and BC = EF.
(c) In ΔABC and ΔQRP, AB = QR, ∠B = ∠R and ∠C = ∠P.
(d) In ΔABC and ΔPQR, AB = PQ, AC = PR and BC = QR.
(e) In ΔABC and ΔPQR, BC = QR, ∠A = 90°, ∠C = ∠R = 40° and ∠Q = 50°.
Answer:
(a) Congruent by SAS (Side-Angle-Side). The included angle is equal.
(b) Congruent by RHS (Right angle-Hypotenuse-Side). AC and DF are hypotenuses.
(c) Congruent by AAS (Angle-Angle-Side). Two angles and a non-included side match.
(d) Congruent by SSS (Side-Side-Side). All three corresponding sides are equal.
(e) Let's find ∠B in ΔABC: 180° - (90° + 40°) = 50°. So ∠B = 50°.
In ΔPQR, ∠P = 180° - (50° + 40°) = 90°.
We have ∠A = ∠P = 90°, ∠B = ∠Q = 50°, ∠C = ∠R = 40°, and BC = QR (hypotenuse).
Congruent by AAS (or ASA).

3. In quadrilateral ABCD, AB = AD and CB = CD. Prove that AC is perpendicular bisector of BD.
Answer:
In ΔABC and ΔADC:
AB = AD (Given)
CB = CD (Given)
AC = AC (Common side)
∴ ΔABC ≅ ΔADC (By SSS congruency criterion)
∴ ∠BAC = ∠DAC (By CPCTC)
Let AC intersect BD at point O. Now in ΔAOB and ΔAOD:
AB = AD (Given)
∠BAO = ∠DAO (Proved above, as O lies on AC)
AO = AO (Common side)
∴ ΔAOB ≅ ΔAOD (By SAS congruency criterion)
∴ BO = DO (By CPCTC), meaning AC bisects BD.
Also, ∠AOB = ∠AOD (By CPCTC).
Since BD is a straight line, ∠AOB + ∠AOD = 180° ⇒ 2∠AOB = 180° ⇒ ∠AOB = 90°.
Thus, AC is the perpendicular bisector of BD.
Hence proved.

4. In the given figure : AB//FD, AC//GE and BD = CE; prove that :
(i) BG = DF
(ii) CF = EG. (HOTS)
Answer:
(i) Since AB // FD and AC // GE, ABC is crossed by parallel lines.
∠B = ∠FDC (Corresponding angles as AB // FD)
∠C = ∠EGB (Wait, AC // GE, so ∠C = ∠GEB, corresponding angles).
Also given BD = CE. Adding DC to both sides: BD + DC = CE + DC ⇒ BC = DE.
In ΔABC and ΔFDE:
∠B = ∠FDE (Corresponding angles)
BC = DE (Proved above)
∠C = ∠FED (Corresponding angles)
∴ ΔABC ≅ ΔFDE (By ASA congruency criterion)
Wait, the points in figure are G, F on AB, AC. It's ΔGBC and ΔFDE or similar.
Looking at the figure carefully: In ΔGBE and ΔFDC:
∠GBE = ∠FDC (Corresponding angles for parallel lines AB and FD with transversal BC)
∠GEB = ∠FCD (Corresponding angles for parallel lines AC and GE with transversal BC)
BD = CE (Given). Adding DE to both sides gives BE = CD.
∴ ΔGBE ≅ ΔFDC (By ASA congruency)
∴ BG = DF (By CPCTC).
Hence proved.

(ii) From the same congruency ΔGBE ≅ ΔFDC (Proved in part i):
EG = CF (By CPCTC).
Hence proved.

5. In a triangle ABC, AB = AC. Show that the altitude AD is median also.
Answer:
Given: ΔABC with AB = AC, and AD is the altitude (so ∠ADB = ∠ADC = 90°).
To show: AD is the median (which means BD = CD).
In right-angled ΔADB and ΔADC:
Hypotenuse AB = Hypotenuse AC (Given)
Side AD = Side AD (Common side)
∠ADB = ∠ADC = 90° (Given, AD is altitude)
∴ ΔADB ≅ ΔADC (By RHS congruency criterion)
∴ BD = CD (By CPCTC)
Since BD = CD, D is the mid-point of BC, making AD the median.
Hence proved.

6. In the following figure, BL = CM. Prove that AD is a median of ΔABC.
Answer:
The figure shows ΔABC with an interior line AD meeting BC at D. Perpendiculars BL and CM are drawn to line AD (or its extension).
In ΔBLD and ΔCMD:
∠BLD = ∠CMD = 90° (Given by perpendicular markings)
∠BDL = ∠CDM (Vertically opposite angles)
BL = CM (Given)
∴ ΔBLD ≅ ΔCMD (By AAS congruency criterion)
∴ BD = CD (By CPCTC)
Since BD = CD, D is the mid-point of BC.
Therefore, AD is a median of ΔABC.
Hence proved.

7. 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:
(i) In right-angled ΔADB and ΔADC:
Hypotenuse AB = Hypotenuse AC (Given)
Side AD = Side AD (Common side)
∠ADB = ∠ADC = 90° (Given, AD ⊥ BC)
∴ ΔADB ≅ ΔADC (By RHS congruency criterion)
∴ BD = CD (By CPCTC).
Hence proved.

(ii) Since BE is the angle bisector of ∠B, ∠FBE = ∠DBE.
In ΔFBE and ΔDBE:
∠BFE = ∠BDE = 90° (Given EF ⊥ AB and AD ⊥ BC)
∠FBE = ∠DBE (Proved above)
BE = BE (Common hypotenuse)
∴ ΔFBE ≅ ΔDBE (By AAS congruency criterion)
∴ EF = ED (By CPCTC).
Hence proved.

8. 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.
Answer:
Let CD intersect AB at point O.
In ΔAOD and ΔBOC:
∠DAO = ∠CBO = 90° (Given, AD and BC are perpendiculars to AB)
∠AOD = ∠BOC (Vertically opposite angles)
AD = BC (Given equal perpendiculars)
∴ ΔAOD ≅ ΔBOC (By AAS congruency criterion)
∴ AO = BO (By CPCTC)
Since AO = BO, the line CD bisects AB at O.
Hence proved.

9. In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that : (HOTS)
(i) BO = CO
(ii) AO bisects angle BAC.
Answer:
(i) In ΔABC, AB = AC, so ∠ABC = ∠ACB (Angles opposite to equal sides are equal).
Since OB and OC are bisectors of ∠B and ∠C:
∠OBC = 1/2 ∠ABC
∠OCB = 1/2 ∠ACB
Therefore, ∠OBC = ∠OCB.
In ΔOBC, sides opposite to equal angles are equal, so BO = CO.
Hence proved.

(ii) In ΔABO and ΔACO:
AB = AC (Given)
BO = CO (Proved in part i)
AO = AO (Common side)
∴ ΔABO ≅ ΔACO (By SSS congruency criterion)
∴ ∠BAO = ∠CAO (By CPCTC)
Thus, AO bisects angle BAC.
Hence proved.

10. In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°. Prove that AD = FC.
Answer:
In right-angled ΔABC and ΔFED:
AB = EF (Given)
BC = DE (Given)
∠B = ∠E = 90° (Given)
∴ ΔABC ≅ ΔFED (By SAS congruency criterion)
∴ AC = FD (By CPCTC)
Now we need to prove AD = FC. From the figure, AD and FC are not direct sides of these triangles, but let's check the segments.
Wait, looking at the figure, points C and D are on a line segment. The base is B-C-D-E.
Given BC = DE. Add CD to both sides:
BC + CD = DE + CD
⇒ BD = CE.
In ΔABD and ΔFEC:
AB = EF (Given)
∠B = ∠E = 90° (Given)
BD = CE (Proved above)
∴ ΔABD ≅ ΔFEC (By SAS congruency criterion)
∴ AD = FC (By CPCTC).
Hence proved.

11. A point O is taken inside a rhombus ABCD such that its distances from the vertices B and D are equal. Show that AOC is a straight line.
Answer:
Given: Rhombus ABCD where AB = BC = CD = DA. Point O inside such that OB = OD.
In ΔABO and ΔADO:
AB = AD (Sides of rhombus)
OB = OD (Given)
AO = AO (Common side)
∴ ΔABO ≅ ΔADO (By SSS congruency criterion)
∴ ∠BAO = ∠DAO (By CPCTC)
This means AO lies on the angle bisector of ∠A.
Similarly, in ΔCBO and ΔCDO:
CB = CD (Sides of rhombus)
OB = OD (Given)
CO = CO (Common side)
∴ ΔCBO ≅ ΔCDO (By SSS congruency)
∴ ∠BCO = ∠DCO (By CPCTC)
This means CO lies on the angle bisector of ∠C.
In a rhombus, the diagonal AC is the angle bisector of both ∠A and ∠C.
Since AO and CO both lie on the bisectors of ∠A and ∠C respectively, they must lie on the diagonal AC.
Therefore, AOC is a straight line (it is the diagonal of the rhombus).
Hence proved.

12. In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ. Prove that ΔAPX ≅ ΔCQY.
Answer:
Since ABCD is a rectangle, opposite sides are equal: AD = BC and CD = AB. ∠D = ∠B = 90°.
Since X and Y are midpoints of DC and AB:
DX = CX = DC/2
AY = BY = AB/2
Since DC = AB, we get DX = CX = AY = BY.
Given DP = BQ.
Since AD = BC, and AD = AP + DP, BC = CQ + BQ.
AP = AD - DP
CQ = BC - BQ
Since AD = BC and DP = BQ, it follows that AP = CQ.
Wait, the points in figure: X is midpoint of top DC. Y is midpoint of bottom AB. P is on AD. Q is on BC. DP=BQ.
To prove ΔAPX ≅ ΔCQY, we need more parts. Let's look at ΔDPX and ΔBQY:
DP = BQ (Given)
∠D = ∠B = 90° (Angles of a rectangle)
DX = BY (Halves of equal opposite sides)
∴ ΔDPX ≅ ΔBQY (By SAS congruency)
∴ PX = QY (By CPCTC).
Now in ΔAPX and ΔCQY:
AP = CQ (Proved above: AD - DP = BC - BQ)
PX = QY (Proved above)
AX = CY (Since ΔADX ≅ ΔCBY by SAS: AD=CB, ∠D=∠B, DX=BY ⇒ AX=CY)
∴ ΔAPX ≅ ΔCQY (By SSS congruency criterion).
Hence proved.

Case-Study Based Question
Mrs. Sharmila, a mathematics teacher of class 9 started a very important geometry chapter on Congruency in triangles and decided to test the previous concept by setting a geometry problem on the blackboard. In ΔABC, AD is drawn as perpendicular bisector of BC. It is given that AD = 9 cm and BC = 24 cm.
Based on the above information answer the following:
(i) Prove ΔABD ≅ ΔACD
(ii) Assign a special name to ΔABC.
(iii) Calculate perimeter of ΔABC.
(iv) Calculate Area of ΔABC.
Answer:
(i) In ΔABD and ΔACD:
BD = CD (Since AD is the perpendicular bisector of BC)
∠ADB = ∠ADC = 90° (Since AD is perpendicular to BC)
AD = AD (Common side)
∴ ΔABD ≅ ΔACD (By SAS congruency criterion)
Hence proved.

(ii) Since ΔABD ≅ ΔACD, we get AB = AC (By CPCTC).
A triangle with two equal sides is called an isosceles triangle.
So, ΔABC is an isosceles triangle.

(iii) AD is the bisector of BC, so BD = BC / 2 = 24 / 2 = 12 cm.
In right-angled ΔABD, using Pythagoras Theorem:
AB² = AD² + BD²
AB² = 9² + 12²
AB² = 81 + 144 = 225
AB = √225 = 15 cm.
Since AB = AC, AC is also 15 cm.
Perimeter of ΔABC = AB + AC + BC
Perimeter = 15 + 15 + 24 = 54 cm.

(iv) Area of ΔABC = (1/2) × base × height
Area = (1/2) × BC × AD
Area = (1/2) × 24 cm × 9 cm
Area = 12 × 9 = 108 cm².