Question-1
Solution:
Given: In the quadrilateral ACBD, AC = AD and AB bisects ∠ A.
To prove: Δ ABC ≅ Δ ABD | Given
Proof: In Δ ABC and Δ ABD,
AB = BA | Common
AC = AD | Given AB bisects ∠ A
∠ CAB = ∠ DAB | SAS Rule
∴ Δ ABC ≅ Δ ABD | CPCT
∴ BC = BD
Question-2
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC
Solution:
Given: ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA.
To Prove: (i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC
Proof: (i) In Δ ABD and Δ BAC,
AD = BC [Given]
AB = BA [AB is common]
ÐDAB = ∠CBA [Given]
∴ ΔABD ≅ ΔBAC [SAS criterion]
(ii) In ΔABD ≅ ΔBAC
∠DAB = ∠CBA
∴ BD = AC [CPCT]
(iii) ΔABD = ΔBAC
AD = BC [Given]
∴ ∠ABD = ∠BAC. [CPCT]
Question-3
Solution:
Given: AD and BC are equal perpendiculars to a line segment AB.
To Prove: CD bisects AB.
Proof: In Δ OAD and Δ OBC
AD = BC | Given
∠ OAD = ∠ OBC | Each = 90°
∠ AOD = ∠ BOC | Vertically Opposite Angles
∴ ∠ ODA = ∠ OCB | AAS Rule
OA = OB | CPCT
\CD bisects AB.
Question-4
Solution:
Given: l and m are two parallel lines intersected by another pair of parallel lines p and q.
To Prove: Δ ABC ≅ Δ CDA.
Proof: AB || DC
AD || BC
∴Quadrilateral ABCD is a parallelogram. |A quadrilateral is a parallelogram if both
|the pairs of opposite sides are parallel ∴ BC = AD ...(1) | Opposite sides of a parallelogram are equal
AB = CD ...(2) | Opposite sides of a parallelogram are equal
And ∠ ABC = ∠ CDA ...(3) | Opposite angles of a parallelogram are equal
In Δ ABC and Δ CDA,
AB = CD | From (2)
BC = DA | From (1)
∠ ABC = ∠ CDA | From (3)
Δ ABC ≅ Δ CDA. | SAS Rule
Question-5
(i) Δ APB ≅ Δ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.
Solution:
Given: Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A.
To Prove: (i) Δ APB ≅ Δ AQB
(ii) BP = BQ Or
B is equidistant from the arms of ∠ A.
Proof: (i) In Δ APB and Δ AQB,
∠ BAP = ∠ BAQ | Since, QAP is the bisector of ∠ A
AB = AB | Common
∠ BPA = ∠ BQA | Each = 90°
(Since BP and BQ are perpendiculars from B to the arms of ∠ A)
∴ Δ APB ≅ Δ AQB | AAS Rule
(ii) Δ APB ≅ Δ AQB | Proved in (i) above
BP = BQ. | CPCT
Question-6
In figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.
Solution:
Given: In figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC.
To Prove: BC = DE.
Proof: In Δ ABC and Δ ADE,
AB = AD | Given
AC = AE | Given
∠ BAD = ∠ EAC | Given
⇒ ∠ BAD + ∠ DAC = ∠ DAC + ∠ EAC | Adding ∠ DAC to both sides
⇒ ∠ BAC = ∠ DAE
∴ Δ ABC ≅ Δ ADE | SAS Rule
∴ BC = DE. | CPCT
Question-7
(i) Δ DAP ≅ Δ EBP
(ii) AD = BE
Solution:
Given: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB.
To Prove : (i) Δ DAP ≅ Δ EBP
(ii) AD = BE.
Proof: (i) In Δ DAP and Δ EBP,
AP = BP | Since P is the midpoint of the line segment AB
∠ DAP = ∠ EBP | Given
∠ EPA = ∠ DPB | Given
⇒ ∠ EPA + ∠ EPD = ∠ EPD + ∠ DPB |Adding ∠ EPD to both sides
⇒ ∠ APD = ∠ BPE | ASA Rule
∴ Δ DAP ≅ Δ EBP
(ii) Δ DAP ≅ Δ EBP | From (1) above
∴ AD = BE. | CPCT
Question-8
In the right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that:(i) Δ AMC ≅ Δ BMD
(ii) ∠ DBC is a right angle
(iii) Δ DBC = Δ ACB
(iv) CM= AB
Solution:
Given: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B.
To Prove: (i) Δ AMC ≅ Δ BMD
(ii) ∠ DBC is a right angle
(iii) Δ DBC ≅ Δ ACB
(iv) CM = AB.
Proof : (i) In Δ AMC and Δ BMD,
AM = BM | Since M is the mid-point of the hypotenuse AB
CM = DM | Given
∠ AMC = ∠ BMD | Vertically Opposite Angles
∴ Δ AMC ≅ Δ BMD. | SAS Rule
(ii) Δ AMC ≅ Δ BMD | From (i) above
∴ ∠ ACM = ∠ BDM | CPCT
But these are alternate interior angles and they are equal
AC || BD
Now, AC || BD and a transversal BC intersects them
∴ ∠ DBC+ ∠ ACB = 180°
| The sum of the consecutive interior angles on the same side of the transversal is 180° ⇒ ∠ DBC + 90° = 180° | Since ∠ ACB = 90° (given)
⇒ ∠ DBC = 180° - 90° = 90°
⇒ ∠ DBC is a right angle.
(iii) In Δ DBC and Δ ACB,
∠ DBC = ∠ ACB (each = 90°) | Proved in (ii) above
BC = CB | Common
∴ Δ AMC = Δ BMD | Proved in (i) above
AC = BD | CPCT
Δ DBC ≅ Δ ACB. | SAS Rule
(iv) Δ DBC ≅ Δ ACB | Proved in (iii) above
DC = AB | CPCT
⇒ 2CM = AB | ∴ DM = CM= DC
⇒ CM = AB.
Question-9
(i) OB = OC (ii) AO bisects ∠ A.
Solution:
Given: In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O.
To Prove: (i) OB = OC
(ii) AO bisects A
Proof: (i) AB = AC | Given
∴ ∠ B = ∠ C | Angles opposite to equal sides of a triangle are equal
∴∠ B = ∠ C |Q BO and CO are the bisectors of ∠ B and ∠ C respectively
∴ ∠ OBC = ∠ OCB
∴ OB = OC | Sides opposite to equal angles of a triangle are equal
(ii) In Δ OAB and Δ OAC, | Given
AB = AC | Given
OB = OC | Proved in (i) above
∠ B = ∠ C | Angles opposite to equal sides of a triangle are equal
∴∠ B = ∠ C
∴ ∠ ABO = ∠ ACO
∴ Δ OAB @ D OAC | Since BO and CO are the bisectors of ∠B and ∠C respectively
∴ ∠ OAB = ∠ OAC | By SAS Rule
∴ AO bisects ∠ A | CPCT
Question-10
In Δ ABC, AD is the perpendicular bisector of BC (see figure). Show that D ABC is an isosceles triangle in which AB = AC.
Solution:
Given: In Δ ABC, AD is the perpendicular bisector of BC.
To Prove: Δ ABC is an isosceles triangle in which AB = AC.
Proof: In Δ ADB and Δ ADC,
∠ ADB = ∠ ADC | Each = 90°
DB = DC | Since AD is the perpendicular bisector of BC
AD = AD | Common
∴ Δ ADB = Δ ADC | By SAS Rule
∴ AB = AC | CPCT
∴ Δ ABC is an isosceles triangle in which AB = AC.
Question-11
Solution:
Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively.
To Prove: BE = CF.
Proof: ABC is an isosceles triangle
∴ AB = AC
∴ ∠ ABC = ∠ ACB ...(1)| Angles opposite to equal sides of a triangle are equal
In Δ BEC and Δ CFB,
Ð BEC = ∠ CFB | Each = 90°
BC = CB | Common
∠ ECB = ∠ FBC | From (1)
∴ Δ BEC ≅ Δ CFB | By ASA Rule
∴ BE = CF. | CPCT
Question-12
(i) Δ ABE ≅ Δ ACF
(ii) AB = AC, i.e., Δ ABC is an isosceles triangle.
Solution:
Given: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal.
To Prove: (i) Δ ABE ≅ Δ ACF
(ii) AB = AC, i.e., Δ ABC is an isosceles triangle.
Proof: (i) In Δ ABE and Δ ACF
BE = CF | Given
∠ BAE = ∠ CAF | Common
Ð AEB = ∠ AFC | Each = 90°
∴ Δ ABE ≅ Δ ACF | By AAS Rule
(ii) Δ ABE ≅ Δ ACF | Proved in (i) above
∴ AB = AC | CPCT
∴ Δ ABC is an isosceles triangle.
Question-13
Solution:
Given: ABC and DBC are two isosceles triangles on the same base BC.
To Prove: ∠ ABD = ∠ ACD
Proof: ABC is an isosceles triangle on the base BC
∴ ∠ ABC = ∠ ACB …..(1)
DBC is an isosceles triangle on the base BC
∴ ∠ DBC = ∠ DCB …..(2)
Adding the corresponding sides of (1) and (2), we get
∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB
⇒ ∠ ABD = ∠ ACD.
Question-14
Solution:
Given: Δ ABC is an isosceles triangle in which AB = AC.
Side BA is produced to D such that AD = AB.
To Prove: ∠ BCD is a right angle.
Proof: ABC is an isosceles triangle
⇒ ∠ ABC = ∠ ACB ...(1)
⇒ AB = AC and AD = AB
⇒ AC = AD
∴ In Δ ACD,
∠ CDA = ∠ ACD | Angles opposite to equal sides of a triangle are equal
⇒ ∠ CDB = ∠ ACD ...(2)
Adding the corresponding sides of (1) and (2), we get
∠ ABC + ∠ CDB = ∠ ACB + ∠ ACD
∠ ABC + ∠ CDB = ∠ BCD …(3)
In Δ BCD,
∠ BCD + ∠ DBC + ∠ CDB = 180° | Sum of all the angles of a triangle is 180°
⇒ ∠ BCD + ∠ ABC + ∠ CDB = 180°
⇒ ∠ BCD + ∠ BCD = 180° | Using (3)
⇒ 2∠ BCD = 180°
Þ ∠ BCD = 90°
⇒ ∠ BCD is a right angle.
Question-15
Solution:
In Δ ABC, AB = AC
∴ ∠ B = ∠ C ...(1) | Angles opposite to equal sides of a triangle are equal
In Δ ABC,
∠ A + ∠ B + ∠ C = 180° | Sum of all the angles of a triangle is 180°
⇒ 90° + ∠ B + ∠ C = 180° | Since ∠ A = 90° (given)
⇒ ∠ B + ∠ C = 90° ...(2)
From (1) and (2), we get
∠ B = ∠ C = 45°.
Question-16
Solution:
Given: An equilateral triangle ABC.
To Prove : ∠ A = ∠ B = ∠ C = 60°.
Proof: ABC is an equilateral triangle
⇒ AB = BC = CA ...(1)
⇒ AB = BC
⇒ ∠ A = ∠ C ...(2) | Angles opposite to equal sides of a triangle are equal
⇒ BC = CA
∴ ∠ A = ∠ B ...(3) | Angles opposite to equal sides of a triangle are equal
From (2) and (3), we obtain
∠ A = ∠ B = ∠ C ... (4)
In Δ ABC,
∠ A + ∠ B + ∠ C = 180^{0} … (5) | sum of all the angles of a triangle is 180^{0 }
From (4) and (5), we get
∠ A = ∠ B = ∠ C = 60^{0}
Question-17
(i) Δ ABD ≅ Δ ACD
(ii)Δ ABP ≅ Δ ACP
(iii) AP bisects ∠ A as well as ∠ D
(iv) AP is the perpendicular bisector of BC.
Solution:
Given: Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.
To Prove: (i) Δ ABD ≅ Δ ACD
(ii) Δ ABP ≅ Δ ACP
(iii) AP bisects ∠ A as well as ∠ D
(iv) AP is the perpendicular bisector of BC.
Proof: (i) In Δ ABD and Δ ACD,
AB = AC …..(1) | Δ ABC is an isosceles triangle
BD = CD ...(2) | Δ DBC is an isosceles triangle
AD = AD ...(3) | Common
∴ Δ ABD @ Δ ACD | SSS Rule
(ii) In Δ ABP and Δ ACP,
AB = AC ...(4) | From (i)
∠ ABP = ∠ ACP ...(5) | AB = AC( from (1)
∴ ÐABP = Ð ACP |Angles opposite to equal sides of a triangle are equal
∴ Δ ABP ≅ Δ ACP | Proved in (i) above
∴ ∠ BAP = ∠ CAP ...(6) | CPCT
In view of (4), (5) and (6)
Δ ABP ≅ Δ ACP
(iii) Δ ABP ≅ Δ ACP
∴ ∠ BAP = ∠ CAP
⇒ AP bisects ∠ A
In Δ BDP and Δ CDP,
BD = CD ...(7) | From (2)
DP = DP ...(8) | Common
∴ Δ ABP = Δ ACP | Proved in (ii) above
\ BP = CP ...(9) | CPCT
In view of (7), (8) and (9),
Δ BDP = Δ CDP
∴ ∠ BDP = ∠ CDP
⇒ DP bisects ∠ D | SSS Rule
⇒ AP bisects ∠ D | CPCT
(iv) Δ BDP = Δ CDP | Proved in (iii) above
\ BP = CP ...(10) | CPCT
∠ BPD = ∠ CPD | CPCT
But ∠ BPD + ∠ CPD = 180° | Linear Pair Axiom
∴ ∠ BPD = ∠ CPD = 90° …..(11)
In view of (10) and (11),
AP is the perpendicular bisector of BC.
Question-18
(i) AD bisects BC
(ii) AD bisects ∠ A.
Solution:
Given: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To Prove: (i) AD bisects BC
(ii) AD bisects ∠ A.
Proof: (i) In right Δ ADB and right Δ ADC,
Hyp. AB = Hyp. AC | Given
Side AD = Side AD | Common
∴ Δ ADB ≅ Δ ADC | RHS Rule
∴ BD = CD | CPCT
⇒ AD bisects BC.
(ii) Δ ADB ≅ Δ ADC | Proved in (i) above
∴ ∠ BAD = ∠ CAD | CPCT
⇒ AD bisects ∠ A.
Question-19
(i) Δ ABM ≅ Δ PQN
(ii) Δ ABC ≅ Δ PQR
Solution:
Given: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR.
To Prove: (i) Δ ABM ≅ Δ PQN
(ii) Δ ABC ≅ Δ PQR.
Proof: (i) In Δ ABM and Δ PQN, ...(1)
AB = PQ ...(2) | Given
AM = PN | Given
BC = QR | M and N are the mid-points of BC and QR respectively
⇒ 2BM = 2QN
⇒ BM = QN ….(3)
In view of (1), (2) and (3),
Δ ABM ≅ Δ PQN | SSS Rule
(ii) Δ ABM ≅ Δ PQN | Proved in (1) above
⇒ ∠ ABM = ∠ PQN | CPCT
⇒ ∠ ABC = ∠ PQR …… (4)
In Δ ABC and Δ PQR,
AB = PQ | Given
BC = QR | Given
∠ ABC = ∠ PQR | From (4)
∴ Δ ABC ≅ ∠ PQR | SAS Rule
Question-20
Solution:
Given: BE and CF are two equal altitudes of a triangle ABC.
To Prove: Δ ABC is isosceles.
Proof: In right Δ BEC and right Δ CFB,
Side BE = Side CF | Given
Hyp. BC = Hyp. CB | Common
∴ Δ BEC ≅ Δ CFB | RHS Rule
∠ BCE = ∠ CBF | CPCT
AB = AC | Sides opposite to equal angles of a triangle are equal
Δ ABC is an isosceles.
Question-21
Solution:
Given: ABC is an isosceles triangle with AB = AC.
To Prove: ∠ B = ∠ C
Construction: Draw AP ⊥ BC
Proof : In right triangle APB and right triangle APC,
Hyp. AB = Hyp. AC | Given
Side AP = Side AP | Common
∴ Δ APB ≅ Δ APC | RHS Rule
∠ ABP = ∠ ACP | CPCT
⇒ ∠ B = ∠ C.
Question-22
Solution:
Let ABC be a right angled triangle in which ∠ B = 90°.
Then, ∠ A + ∠ C = 90° | Sum of all the angles of a triangle is 180°
∴ ∠ B = ∠ A + ∠ C
∴ ∠ B > ∠ A
and ∠ B > ∠ C
∴ AC > BC | ∴ Side opposite to greater angle is longer and AC > AB
∴ AC is the longest side, i.e., hypotenuse is the longest side.
Question-23
Solution:
Given: Sides AB and AC of Δ ABC are extended to points P and Q respectively. Also,
∠ PBC < ∠ QCB.
To Prove: AC >AB.
Proof: ∠ PBC < ∠ QCB | Given
⇒ - ∠ PBC > - ∠ QCB
⇒ 180° - ∠ PBC > 180° - ∠ QCB
⇒ ∠ ABC > ∠ ACB
∴ AC > AB |Since the side opposite to the greater angle is longer
Question-24
In figure, ∠ B < ∠ A and ∠ C < ∠ D, Show that AD < BC.
Solution:
Given: In figure, ∠ B < ∠ A and ∠ C < ∠ D.
To Prove: AD < BC | Given
Proof: ∠ B < ∠ A
∴ ∠ A > ∠ B
OB > OA ...(1) | Side opposite to greater angle is longer
∠ C < ∠ D | Given
∠ D > ∠ C
OC > OD ...(2) | Side opposite to greater angle is longer
From (1) and (2), we get
OB + OC > OA + OD
⇒ BC > AD
⇒ AD < BC.
Question-25
Solution:
Given: AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.
To Prove: ∠ A > ∠ C and ∠ B >∠ D
Construction: Join AC.
Proof: In Δ ABC,
AB < BC | Given, AB is the smallest side of quadrilateral ABCD
⇒ BC > AB
∴ ∠ BAC > ∠ BCA ...(1) | Angle opposite to longer side is greater
In Δ ACD, CD > AD | Given, CD is the longest side of quadrilateral ABCD
∠ CAD > ∠ ACD | Angle opposite to longer side is greater ….(2)
From (1) and (2), we obtain
∠ BAC + ∠ CAD > ∠ BCA + ∠ ACD
⇒ ∠ A > ∠ C
Similarly, joining B to D, we can prove that ∠ B > ∠ D.
Question-26
In figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.
Solution:
Given: In figure, PR > PQ and PS bisects ∠ QPR.
To Prove: ∠ PSR > ∠ PSQ
Proof: In Δ PQR,
PR > PQ (Given)
∴ ∠ PQR > ∠ PRQ ...(1) ( ∵ Angle opposite to longer side is greater)
∴ PS is the bisector of ∠ QPR
∴ ∠ QPS = ∠ RPS ...(2)
In Δ PQS,
∠ PQS + ∠ QPS + ∠ PSQ = 180° ...(3) (∵ The sum of the three angles of a triangle is 180°)
In Δ PRS,
∠ PRS + ∠ SPR + ∠ PSR = 180° ...(4)
Sum of the three angles of triangle is 180° from (3) and (4),
∠ PQR + ∠ QPS + ∠ PSQ = ∠ PRS + ∠ SPR + ∠ PSR
⇒ ∠ PQR + ∠ PSQ = ∠ PRS + ∠ PSR (from (2) )
⇒ ∠ PRS + ∠ PSR = ∠ PQR + ∠ PSQ
⇒ ∠ PRS + ∠ PSR > ∠ PRQ + ∠ PSQ (From (1))
⇒ ∠ PRQ + ∠ PSR > ∠ PRS + ∠ PSQ ( ∵ ∠ PRQ = ∠ PRS)
⇒ ∠ PSR > ∠ PSQ
Question-27
Solution:
Given: l is a line and P is a point not lying on l. PM ⊥ l. N is any point on l other than M.
To Prove: PM < PN.
Proof: In Δ PMN,
∠ M = 90°
⇒ ∠ N is an acute angle. | Angle sum property of a triangle
⇒ ∠ M > ∠ N
⇒ PN > PM | Side opposite to greater angle is greater
∴ PM < PN.
Question-28
Solution:
Construct the perpendicular bisectors of two sides of Δ ABC. Their point of intersection is the required point.
Question-29
Solution:
The angle bisector between of any two angles of the triangle. Their point of intersection is the required point which is the incentre. Incentre is equidistant from all the sides of triangle.
Question-30
Solution:
Draw the perpendicular bisectors of AB and AC. Their point of intersection is the required point.
Question-31
(i) (ii) |
Solution:
(i) Number of triangles = 25 + 25 + 25 + 25 + 25 + 25 = 150
(i) |
(ii) Number of triangles = 24 × 12 = 288.
(ii) |
Figure (ii) has more triangles.