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Given: In quadrilateral ACBD, AC = AD and AB bisects ∠A.
To Prove: ∆ABC ≅ ∆ABD.
Proof: In ∆ABC and ∆ABD,
AC = AD | Given
AB = AB | Common
∠CAB = ∠DAB
| ∵ AB bisects ∠A
∴ ∠ABC ≅ ∠ABD | SAS Rule
∴ BC = BD | C.P.C.T,
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Given: ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA.
To Prove: (i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Proof: (i) In ∆ABD and ∆BAC,
AD = BC | Given
AB = BA | Common
∠DAB = ∠CBA | Given
∴ ∆ABD ≅ ∠BAC | SAS Rule
(ii) ∵ ∆ABD ≅ ∆BAC | Proved in (i)
∴ BD = AC | C.P.C.T.
(iii) ∵ ∆ABD ≅ ∠BAC | Proved in (i)
∴ ∠ABD = ∠BAC. | C.P.C.T.
Given: AD and BC are equal perpendiculars to a line segment AB.
To Prove: CD bisects AB.
Proof: In ∆O AD and ∆OBC
AD = BC | Given
∠OAD = ∠OBC | Each = 90°
∠AOD = ∠BOC
| Vertically Opposite Angles
∴ ∠OAD ≅ ∆OBC | AAS Rule
∴ OA = OB | C.P.C.T.
∴ CD bisects AB.
Given: I and m are two parallel lines intersected by another pair of parallel lines p and q.
To Prove: ∆ABC ≅ ∆CDA.
Proof: ∵ AB || DC
and 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 ||gm are equal
AB = CD ...(2)
| Opposite sides of a ||gm are equal
and ∠ABC = ∠CDA ...(3)
| Opposite angles of a ||gm are equal
In ∆ABC and ∆CDA,
AB = CD | From (2)
BC = DA | From (1)
∠ABC = ∠CDA | From (3)
∴ ∆ABC ≅ ∆CDA. | SAS Rule
Line I is the bisector of an angle ∠A and B is any point on I. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
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) ∆APB ≅ ∆AQB
(ii) BP = BQ
Or
B is equidistant from the arms of ∠A.
Proof: (i) In ∆APB and ∆AQB,
∠BAP = ∠BAQ
| ∵ l is the bisector of ∠A
AB = AB | Common
∠BPA = ∠BQA | Each = 90°
| ∵ BP and BQ are perpendiculars from B to the arms of ∠A
∴ ∆APB ≅ ∆AQB | AAS Rule
(ii) ∵ ∆APB ≅ ∆AQB
| Proved in (i) above
∴ BP = BQ. | C.P.C.T.
Given: In figure, AC = AE, AB = AD and ∠BAD = ∠EAC.
To Prove: BC = DE.
Proof: In ∆ABC and ∆ADE,
AB = AD | Given
AC = AE | Given
∠BAD = ∠EAC | Given
⇒ ∠BAD + ∠DAC = ∠DAC + ∠EAC
| Adding ∠DAC to both sides
⇒ ∠BAC = ∠DAE
∴ ∆ABC ≅ ∆ADE | SAS Rule
∴ BC = DE. | C.P.C.T.
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE.
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) ∆DAP ≅ ∆EBP
(ii) AD = BE.
Proof: (i) In ∆DAP and ∆EBP,
AP = BP
| ∵ P is the mid-point of the line segment AB
∠DAP = ∠EBP | Given
∠EPA = ∠DPB | Given
⇒ ∠EPA + ∠EPD = ∠EPD + ∠DPB
| Adding ∠EPD to both sides
⇒ ∠APD = ∠BPE
∴ ∠DAP ≅ ∠EBP | ASA Rule
(ii) ∵ ∆DAP ≅ AEBP | From (i) above
∴ AD = BE. | C.P.C.T.
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB
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) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∠DBC ≅ ∆ACB
(i) In ∆AMC and ∆BMD,
AM = BM
| ∵ M is the mid-point of the hypotenuse AB
CM = DM | Given
∠AMC = ∠BMD
| Vertically Opposite Angles
∴ ∆AMC ≅ ∆BMD. | SAS Rule
(ii) ∵ ∆AMC ≅ ∆BMD
| From (i) above
∠ACM = ∠BDM | C.P.C.T.
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 a transversal is
180°
⇒ ∠DBC + 90° = 180°
| ∵ ∠ACB = 90° (given)
⇒ ∠DBC = 180° - 90° = 90°
⇒ ∠DBC is a right angle.
(iii) In ∆DBC and ∆ACB,
∠DBC = ∠ACB (each = 90°)
| Proved in (ii) above
BC = CB | Common
∵ ∆AMC ≅ ∆BMD | Proved in (i) above
∴ AC = BD | C.P.C.T.
∴ ∆DBC ≅ ∆ACB. | SAS Rule
(iv) ∵ ∆DBC ≅ ∆ACB
| Proved in (iii) above
∴ DC = AB | C.P.C.T.
2CM = AB
In ∆QPR and ∆PQS,
QR = PS | Given
∠RQP = ∠SPQ | Given
PQ = PQ | Common
∴ ∆QPR ≅ ∆PQS | SAS Axiom
∴ PR = QS | C.P.C.T.
and ∠QPR = ∠PQS. | C.P.C.T.
In ∆OAP and ∆OBQ,
AP = BQ | Given
∠OAP = ∠OBQ | Each = 90°
∠AOP = ∠BOQ
| Vertically Opposite Angles
∴ ∆OAP ≅ ∆OBQ | AAS Axiom
∴ OA = OB | C.P.C.T.
and OP = OQ | C.P.C.T.
⇒ O is the mid-point of line segments AB and PQ
In ∆ABC and ∆ADC,
∠BAC = ∠DAC | Given
∠ACB = ∠ACD | Given
AC = AC | Common
∴ ∆ABC ≅ ∆ADC | ASA Axiom
∴ AB = AD | C.P.C.T.
and CB = CD. | C.P.C.T.
AB is a line-segment. AX and BY are two equal line-segments drawn on opposite sides of line AB such that AX || BY. If AB and XY intersect each other at P. Prove that:
(i) ∆APX ≅ ∆BPY
(ii) AB and XY bisect each other at P.
(i) ∵ AX || BY and AB intersects them
∴ ∠PAX = ∠PBY ...(1)
| Alternate Angles
∵ AX || BY and XY intersects them
∴ ∠PXA = ∠PYB ...(2)
| Alternate Angles
In ∆APX and ∆BPY,
∠PAX = ∠PBY | From (1)
∠PXA = ∠PYB | From (2)
AX = BY | Given
∴ ∆APX = ∆BPY | ASA Axiom
(ii) ∵ AP = BP | C.P.C.T.
and PX = PY | C.P.C.T.
⇒ AB and XY bisect each other at P.
Given: ∠QPR = ∠PQR and M and N are respectively points on side QR and PR of ∆PQR, such that QM = PN.
To Prove: OP = OQ, where O is the point of intersection of PM and QN.
Proof: In ∆PNQ and ∆QMP,
PN = QM | Given
PQ = QP | Common
∠QPN = ∠PQM | Given
∴ ∆PNQ ≅ ∆QMP
| SAS congruence rule
∴ ∠PNQ = ∠QMP | CPCT
Again, in ∆PNO and ∆QMO,
PN = QM | Given
∠PON = ∠QOM
| Vertically opposite angles
∠PNO = ∠QMO | Proved above
∴ ∆PNO ≅ ∆QMO
| AAS congruence rule
∴ OP = OQ | CPCT
Given: ABC is an equilateral triangle whose medians are AD, BE and CF.
To Prove: AD = BE = CF
Proof: In ∆ADC and ∆BEC,
AC = BC
DE = EC
| SAS congruence rule
....(1) | CPCT
Similarly, we can prove that
BE = CF ...(2)
and CF = AD ...(3)
From (1), (2) and (3)
AD = BE = CF
Given: In figure,
∠B = ∠E, BD = CE
and ∠1 = ∠2
To Prove: ∆ABC ≅ ∆AED
Proof: ∠1 = ∠2
⇒ ∠1 + ∠DAC = ∠2 + ∠DAC
⇒ ∠BAC = ∠EAD ...(1)
BD = CE
⇒ BD + DC = CE + DC
⇒ BC = ED ...(2)
∠B = ∠E ...(3)
In view of (1), (2) and (3),
∆ABC ≅ ∆AED
| AAS congruence rule
In figure given below, AD is the median of ∆ABC.
BE ⊥ AD, CF ⊥ AD. Prove that BE = CF.
Given: AD is the median of ∆ABC. BE ⊥ AD, CF ⊥ AD.
To Prove: BE = CF
Proof: In ∆DEB and ∆DFC,
DB = DC
| ∵ AD is the median of ∆ABC
∠DEB = ∠DFC | Each = 90°
∠BDE = ∠CDF
| Vertically opposite angles
∴ ∆DEB ≅ ∆DFC
| AAS congruence rule
∴ BE = CF | CPCT
Given:
AB = FE, BC = ED,
AB ⊥ BD and FE ⊥ EC
To Prove: AD = FC
Proof: In ∆ABD and ∆FEC,
AB = FE ...(1) | Given
∠ABD = ∠FEC ...(2)
| Each = 90°
BC = ED | Given
⇒ BC + CD = ED + DC
⇒ BD = EC ...(3)
In view of (1), (2) and (3),
∆ABD ≅ ∆FEC
| SAS congruence rule
∴ AD = FC | CPCT
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In figure, OA = OB and OD = OC. Show that:
(i) ∆AOD ≅ ∆BOC and (ii) AD = BC.
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In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠A.
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
∴ ∠OBC = ∠OCB
| ∵ BO and CO are the bisectors of ∠B and ∠C respectively
∴ OB = OC
| Sides opposite to equal angles of a triangle are equal
(ii) In ∆OAB and ∆OAC,
AB = AC | Given
OB = OC | Proved in (i) above
OA = OA | Common
∴ ∠B = ∠C
| Angles opposite to equal sides of a triangle are equal
∴ ∠ABO = ∠ACO
| ∵ BO and CO are the bisectors of ∠B and ∠C respectively
∴ ∆OAB ≅ ∆OAC | By SAS Rule
∴ ∠OAB = ∠OAC | C.P.C.T.
∴ AO bisects ∠A.
Given: In ∆ ABC, AD is the perpendicular bisector of BC.
To Prove: A ABC is an isosceles triangle in which AB = AC.
Proof: In ∆ ADB and ∆ADC,
∠ADB = ∠ADC | Each = 90° DB = DC
| ∵ AD is the perpendicular bisector of BC
AD = AD | Common
∴ ∆DB ≅ ∆ADC | By SAS Rule
∴ AB = AC | C.P.C.T.
∴ ∆ABC is an isosceles triangle in which AB = AC.
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 ∆BEC and ∆CFB,
∠BEC = ∠CFB | Each = 90°
BC = CB | Common
∠ECB = ∠FBC | From (1)
∴ ∆BEC ≅ ∆CFB | By AAS Rule
∴ BE = CF. | C.P.C.T.
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC, i.e., ∆ABC is an isosceles triangle.
Given: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal.
To Prove: (i) ∆ABE = ∆ACF
(ii) AB = AC, i.e., ∆ABC is an isosceles triangle.
Proof: (i) In ∆ABE and ∆ACF
BE = CF | Given
∠BAE = ∠CAF | Common
∠AEB = ∠AFC | Each = 90°
∴ ∆ABE ≅ ∆ACF | By AAS Rule
(ii) ∆ABE ≅ ∆ACF | Proved in (i) above
∴ AB = AC | C.P.C.T.
∴ ∆ABC is an isosceles triangle.
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.
∴ ∠BC = ∠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.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∆BCD is a right angle.
Given: ∆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 ∆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 ∆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°
⇒ ∠BCD = 90°
⇒ ∠BCD is a right angle.
∵ In ∆ABC,
AB = AC
∴ ∠B = ∠C ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆ABC,
∠A + ∠B + ∠C = 180°
| Sum of all the angles of a triangle is 180°
⇒ 90° + ∠B + ∠C = 180°
| ∵ ∠A = 90° (given)
⇒ ∠B + ∠C = 90° ...(2)
From (1) and (2), we get
∠B = ∠C = 45°.
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 ∆ABC,
∠A + ∠B + ∠C = 180° ...(5)
| Sum of all the angles of a triangle is 180°
Let ∠A = x°. Then, ∠B = ∠C = x°
| From (4)
From (5),
x° + x° + x° = 180°
3x° = 180°
⇒ x° = 60°
⇒ ∠A = ∠B = ∠C = 60°.
∵ ∠A = ∠B
∴ BC = CA ...(1)
| Sides opposite to equal angles of ∆ABC
∵ ∠B = ∠C
∴ CA = AB ...(2)
| Sides opposite to equal angles of ∆ABC
∵ ∠C = ∠A
∴ AB = BC ...(3)
| Sides opposite to equal angles of ∆ABC
From (1), (2) and (3), we have
AB = BC = CA
⇒ ∆ABC is equilateral.
∵ AB = AC
∴ ∠ACB = ∠ABC
| Angles opposite to equal sides of ∆ABC
⇒ 2∠2 = 2∠1
| ∵ CE and BD are the bisectors of ∠C and ∠B respectively
⇒ ∠2 = ∠1
⇒ BP = PC ...(1)
| Sides opposite to equal angles of ∆PBC
In ∆BPE and ∆CPD,
BP = CP | Proved above
∠EBP = ∠DCP | Proved above
∠BPE = ∠CPD
| Vertically Opposite Angles
∴ ∆BPE ≅ ∆CPD | ASA Axiom
PE = PD | C.P.C.T.
⇒ PD = PE ...(2)
Adding (1) and (2), we get
BP + PD = PC + PE
⇒ BD = CE.
ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.
[Hint. Take a point P on AC such that BP bisects ∠B. Join P and D.]
Construction: Take a point P on AC such that BP bisects ∠B. Join P and D.
Proof: In ∆ABC,
∵ BP bisects ∠ABC
In APBC,
∴ ∠PBC = ∠PCB (= ∠C)
∴ PB = PC ...(2)
| Sides opposite to equal angles of∆PBC In ∆APB and ∆DPC,
AB = CD | Given
PB = PC | From (2)
∠ABP = ∠DCP (= ∠C)
∴ ∆APB ≅ ∆DPC | SAS Axiom
∴ ∠BAP = ∠CDP (= ∠A) ...(3)
| C.P.C.T.
and AP = DP ...(4) | C.P.C.T.
In ∆APD,
∵ AP = DP | From (4)
Again from ∆DPC,
∠DPC = π - (∠A + ∠C)
∴ ∠DPA = π - ∠DPC = π - {π - (∠A + ∠C)} = ∠A + ∠C ...(6)
From (5) and (6),
π - A = ∠A + ∠C ⇒ 2∠A + ∠C = π ...(7)
Again,
∠A + ∠B + ∠C = π
| ∵ The sum of three angles of ∆ABC = π ⇒ ∠A + 2∠C + ∠C = π | ∵ ∠B = 2∠C
⇒ ∠A + 3∠C = π ...(8)
Multiplying (7) by 3, we get
6∠A + 3∠C = 3π ...(9)
Subtracting (8) from (9), we get
In ∆OAC and ∆ODB,
OA = OD | Given
OB = OC | Given
∠AOC = ∠DOB
| Vertically Opposite Angles
∴ ∆OAC ≅ ∆ODB | SAS Axiom
∴ AC = BD | C.P.C.T.
Also, ∠OAC = ∠ODB | C.P.C.T.
and ∠OCA = ∠OBD | C.P.C.T.
Thus ∠OAC may not be equal to ∠OBD and therefore, AC may not be parallel to BD
However, if OA = OC, then ∠OAC = ∠OCA
| Angles opposite to equal sides of ∆OAC But ∠OAC = ∠ODB
∴ ∠OCA = ∠ODB
But these angles form a pair of equal alternate angles
∴ AC || BD.
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Given: D and E are points on the base BC of a ∆ABC such that AD = AE and ∠BAD = ∠CAE.
To Prove: AB = AC
Proof: In ∆ADE,
∵ AD = AE | Given
∴ ∠ADE = ∠AED ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆ABD,
Ext. ∠ADE = ∠BAD + ∠ABD ...(2)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
In ∆AEC,
Ext. ∠AED = ∠CAE + ∠ACE .. .(3)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
From (1), (2) and (3),
∠BAD + ∠ABD = ∠CAE + ∠ACE
⇒ ∠ABD = ∠ACE
| ∵ ∠BAD = ∠CAE (Given)
⇒ ∠ABC = ∠ACB
∴ AB = AC
| Sides opposite to equal angles of a triangle are equal
Given: X and Y are two points on equal sides AB and AC of a ∆ABC such that AX = AY.
To Prove: XC = YB
Proof: In ∆ABC,
∵ AB = AC | Given
∴ ∠ABC = ∠ACB ...(1)
| Angles opposite to equal sides of a triangle are equal
Again, AB = AC | Given
AX = AY | Given
Subtracting, we get,
AB - AX = AC - AY
⇒ BX = CY ...(2)
In ∆BXC and ∆CYB,
BX = CY | From (2)
BC = CB | Common
∠XBC = ∠YCB | From (1)
∴ ∆BXC ≅ ∆CYB
| SAS congruence rule
∴ XC = YB | CPCT
Given: A triangle ABC in which AB = AC
To Prove: ∠ABC = ∠ACB
Construction: Draw the bisector AD of A so as to intersect BC at D.
Proof: In ∆ADB and ∆ADC,
AD = AD | Common
AB = AC | Given
∠BAD = ∠CAD
| By Construction
∴ ∆ADB ≅ ∆ADC
| SAS congruence rule
∴ ∠ABD = ∠ACD | CPCT
⇒ ∠ABC = ∠ACB
Yes, the converse is true.
Given: AB = AC, D is the point in the interior of ∆ABC such that ∠DBC = ∠DCB.
To Prove: AD bisects ∠BAC of ∆ABC.
Proof: In ∆DBC,
∵ ∠DBC = ∠DCB ...(1) | Given
∴ DB = DC ...(2)
| Sides opposite to equal angles of a triangle are equal
In ∆ABC,
∵ AB = AC | Given
∴ ∠ABC = ∠ACB
∴ ∠ABC = ∠ACB ...(3)
| Angles opposite to equal sides of a triangle are equal
Subtracting (1) from (3), we get,
∠ABC - ∠DBC = ∠ACB - ∠DCB
⇒ ∠ABD = ∠ACD ...(4)
In ∆ADB and ∆ADC,
AB = AC | Given
DB = DC | Proved in (2)
∠ABD = ∠ACD | Proved in (4)
∴ ∆ADB ≅ ∆ADC
| SAS congruence rule
∴ ∠DAB = ∠DAC | CPCT
⇒ AD bisects ∠BAC of ∆ABC.
In figure, ABCD is a square and ∠DEC is an equilateral triangle. Prove that
(i) ∆ADE ≅ ∆BCE
(ii) AE = BE
(iii) ∠DAE = 15°
Given: ABCD is a square and ∆DEC is an equilateral triangle.
To Prove:
(i) ∆ADE ≅ ∆BCE
(ii) AE = BE
(iii) ∠DAE = 15°
Proof: (i) In ∆ADE and ∆BCE,
AD = BC
DE=CE
∴ ∆ADE ≅ ∆BCE | SAS congruence rule
(ii) ∵ ∆ADE ≅ ∆BCE | Proved in (1)
∴ AE = BE | CPCT
(iii) In ∆DAE,
∵ DE = DA | Given
∴ ∠DAE = ∠DEA ...(1)
| Angles opposite to equal sides of a triangle are equal Also, ∠ADE + ∠DAE + ∠DEA = 180°
| Angle sum property of a triangle
⇒ (∠ADC + ∠EDC) + ∠DAE + ∠DEA = 180°
⇒ (90° + 60°) + ∠DAE + ∠DEA = 180°
⇒ ∠DAE + ∠DEA = 30° ...(2)
From (1) and (2),
∠DAE = 15° = ∠DEA
Given: In ∆ABC, AB = AC, ∠A = 36°. The internal bisector of ∠C meets AB at D.
To Prove: AD = DC
Proof: In ∆ABC,
∵ AB = AC
∴ ∠ABC = ∠ACB ...(1)
| Angles opposite to equal sides of a triangle are equal
Also, ∠BAC + ∠ABC + ∠ACB = 180°
| Angle sum property of a triangle
⇒ 360° + ∠ABC + ∠ACB = 180°
⇒ ∠ABC + ∠ACB = 144° ...(2)
From (1) and (2),
Now, ∵ CD bisects ∠ACB
Again, In ∆ACD,
∠ACD = ∠CAD (= 36°)
∴ AD = DC
| Sides opposite to equal angles of a triangle are equal
Given: AB = BC, AD = EC
To Prove: ∆ABE ≅ ∆CBD
Proof: In ∆ABC,
∵ AB = BC | Given
∴ ∠BAC = ∠BCA ...(1)
| Angles opposite to equal sides of a triangle are equal
AD = EC | Given
⇒ AD + DE = EC + DE
⇒ AE = CD ...(2)
Now, in ∆ABE and ∆CBD,
AE = CD | From (2)
AB = CB | Given
∠BAE = ∠BCD | From (1)
∴ ∆ABE ≅ ∆CBD | SAS congruence rule.
Given: ABC is an isosceles triangle with
AB = AC.
AP ⊥ BC
To Prove: ∠B = ∠C
Proof: In ∆ABC,
∵ AB = AC | Given
∴ ∠ABC = ∠ACB ...(1)
| Angles opposite to equal sides of a triangle are equal
Now, in ∆APB and ∆APC,
AB = AC | Given
∠ABP = ∠ACP | From (1)
∠APB = ∠APC (= 90°) | Given
∴ ∆APB ≅ ∆APC | AAS congruence rule
∴ ∠ABP = ∠ACP | CPCT
⇒ ∠B = ∠C
Given: In an isosceles triangle ABC with AB = AC, BD and CE are two medians.
To Prove: BD = CE
Proof: In ∆ABC,
∵ AB = AC
∴ ∠BC = ∠ACB ...(1)
| Angles opposite to equal sides of a triangle are equal
Also, 
| Halves of equals are equal ⇒ BE = CD ...(2)
| ∵ BD and CE are two medians
Now, in ∆BDC and ∆CEB,
∠BCD = ∠CBE | From (1)
BE = CD | From (2)
BC = CB | Common
∴ ∆BDC ≅ ∆CEB
| SAS congruence rule
∴ BD = CE | CPCT
Given: In figure, ∠x = ∠y and PQ = PR
To Prove: PE = RS
Construction: Join PR
Proof: In ∆PQR,
∵ PQ = QR | Given
∴ ∠QRP = ∠QPR
| Angles opposite to equal sides of a triangle are equal
⇒ ∠ERP = ∠SPR ...(1)
In ∆PER and ∆RSP,
∠ERP = ∠SPR From (1)
∠REP = ∠PSR | Given
PR = RP | Common
∴ ∆PER ≅ ∆RSP
| AAS congruence rule
∴ PE = RS | CPCT
Given: AE bisects ∠DAC and ∠B = ∠C
To Prove: AE || BC
Proof: In ∆ABC,
Ext. ∠DAC = ∠ABC + ∠ACB ...(1)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
⇒ ∠DAC = ∠ACB + ∠ACB
| ∵ ∠B = ∠C (Given)
⇒ ∠DAC = 2∠ACB
⇒ 2∠CAE = 2∠ACB
⇒ ∠CAE = ∠ACB
But these angles form a pair of equal alternate interior angles
∴ AE || BC
Given: A ∆ABC in which the bisector of the vertical angle ∠BAC bisects the base BC, i.e., BD = CD
To Prove: ∆ABC is isosceles
Construction: Produce AD to E such that AD = DE. Join EC.
Proof: In ∆ADB and ∆EDC,
BD = CD | Given
AD = ED | By construction
∠ADB = ∠EDC
| Vertically opposite angles
∴ ∆ADB ≅ ∆EDC
| SAS congruence rule ∴ AB = EC ...(1) | CPCT
and ∠BAD = ∠CED | CPCT
But ∠BAD = ∠CAD | Given
∴ ∠CAD = ∠CED
∴ AC = CE ...(2)
| Sides opposite to equal angles of a triangle are equal
From (1) and (2),
AB = AC
∴ ∆ABC is isosceles.
In ∆ABC,
∵ ∠A = ∠B | Given
∴ ∠CAB = ∠CBA (= 85°)
| Angles opposite to equal sides of a triangle are equal
Also, x + ∠CAB + ∠CBA = 180°
| Angle sum property of a triangle
⇒ x + 85° + 85° = 180°
⇒ x = 10°
In ∆BCD,
∵ BD = CD
∴ x = z = 10°
| Angles opposite to equal sides of a triangle are equal
Also, z + x + y = 180°
| Angle sum property of a triangle
⇒ 10° + 10° + y = 180°
⇒ y = 160°
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(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Given: ∆ABC and ∆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) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Proof: (i) In ∆ABD and ∆ACD,
AB = AC ...(1)
| ∵ ∆ABC is an isosceles triangle
BD = CD ...(2)
| ∵ ADBC is an isosceles triangle
AD = AD ...(3) | Common
∴ ∆ABD ≅ ∆ACD | SSS Rule
(ii) In ∆ABP and ∆ACP,
AB = AC ...(4) | From(1)
∠ABP = ∠ACP ...(5)
| ∵ AB = AC From (1) ∴ ∠ABP = ∠ACP Angles opposite to equal sides of a triangle are
equal
∵ ∆ABD ≅ ∆ACD
| Proved in (i) above
∴ ∠BAP = ∠CAP ...(6) | C.P.C.T.
In view of (4), (5) and (6)
∆ABP ≅ ∆ACP | ASA Rule
(iii) ∵ ∆ABP ≅ ∆ACP
| Proved in (ii) above
∠BAP = ∠CAP | C.P.C.T.
⇒ AP bisects ∠A.
In ∆BDP and ∆CDP,
BD = CD ...(7) | From (2)
DP = DP ...(8) | Common
∵ ∆ABP ≅ ∆ACP
| Proved in (ii) above
∴ BP = CP ...(9) | C.P.C.T.
In view of (7), (8) and (9),
∆BDP ≅ ∆CDP | SSS Rule
∴ ∠BDP = ∠CDP | C.P.C.T.
⇒ DP bisects ∠D
⇒ AP bisects ∠D
(iv) ∵ ∆BDP ≅ ∆CDP
| Proved in (iii) above
∴ BP = CP ...(10) | C.P.C.T.
∠BPD = ∠CPD | C.P.C.T.
But ∠BPD + ∠CPD = 180°
| Linear Pair Axiom
∴ ∠BPD = ∠CPD = 90° ...(11)
In view of (10) and (11),
AP is the perpendicular bisector of BC.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
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 ∆ADB and right ∆ADC,
Hyp. AB = Hyp. AC | Given
Side AD = Side AD | Common
∴ ∆ADB ≅ ∆ADC | RHS Rule
∴ BD = CD | C.P.C.T.
⇒ AD bisects BC.
(ii) ∵ ∆ADB ≅ ∆ADC
| Proved in (i) above
∴ ∠BAD = ∠CAD | C.P.C.T.
⇒ AD bisects ∠A.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of triangle PQR (see figure). Show that:
(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR.
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 ∆PQR.
To Prove: (i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR.
Proof: (i) In ∆ABM and ∆PQN,
AB = PQ ...(1) | Given
AM = PN ...(2) | Given
BC = QR
⇒ 2BM = 2QN
| ∵ M and N are the mid-points of BC and QR respectively
⇒ BM = QN ...(3)
In view of (1), (2) and (3),
∆ABM ≅ ∆PQN | SSS Rule
(ii) ∵ ∆ABM ≅ ∆PQN
| Proved in (1) above
∴ ∠ABM = ∠PQN | C.P.C.T.
⇒ ∠ABC = ∠PQR ...(4)
In ∆ABC and ∆PQR,
AB = PQ | Given
BC = QR | Given
∠ABC = ∠PQR | From (4)
∴ ∆ABC ≅ ∆PQR. | SAS Rule
Given: BE and CF are two equal altitudes of a triangle ABC.
To Prove: ∆ABC is isosceles.
Proof: In right ∆BEC and right ∆CFB,
Side BE = Side CF | Given
Hyp. BC = Hyp. CB | Common
∴ ∆BEC ≅ ∆CFB | RHS Rule
∴ ∠BCE = ∠CBF | C.P.C.T.
∴ AB = AC
| Sides opposite to equal angles of a triangle are equal
∴ ∆ABC is isosceles.
Given: ABC is an isosceles triangle with AB = AC.
To Prove: ∠B = ∠C
Construction: Draw AP π BC
Proof: In right triangle APB and right triangle
APC,
Hyp. AB = Hyp. AC | Given
Side AP = Side AP | Common
∴ ∆APB ≅ ∆APC | RHS Rule
∴ ∠ABP = ∠ACP | C.P.C.T.
⇒ ∠B = ∠C.
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In ∆PQR, ∠P = 100° and ∠R = 60°, which side of the triangle is the longest. Give reasons for your answer.
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In the given figure, if AD is the bisector of ∠BAC then prove that
(i) AB > BD (ii) AC > CD
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In a huge park, people are concentrated at three points (see figure):
A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit. Where should an icecream parlour be set up so that maximum number of persons can approach it?
[Hint: The parlour should be equidistant from A, B and C.]
(ii) Number of triangles = 24 × 12 = 288
Figure (ii) has more triangles. D.
RHS congruence rule.
B.
isoscelesD.
The two altitudes corresponding to two equal sides of a triangle are not equal.Sponsor Area
C.
AB = DE, BC = EF, CA = FDD.
AB < BC < AC
C.
∠B is the smallest angle in triangleD.
∠A = ∠D, ∠B = ∠E, ∠C = ∠FB.
∆CBA ≅ ∆PRQ
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In figure, AB ⊥ AE, BC ⊥ AB, CE = DE and ∠AED = 120°. Find
(a) ∠EDC
(b) ∠DEC
(c) Hence prove that EDC is an equilateral triangle.
Solution not provided.
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