Triangles — Exercise 8(A)

Which of the following pairs of triangles are congruent ?

(a) △ABC and △DEF in which : BC = EF, AC = DF and ∠C = ∠F.

(b) △ABC and △PQR in which : AB = PQ, BC = QR and ∠C = ∠R.

(c) △ABC and △LMN in which : ∠A = ∠L = 90°, AB = LM, ∠C = 40° and ∠M = 50°

(d) △ABC and △DEF in which : ∠B = ∠E = 90°, AC = DF

Answer

(a) In △ABC and △DEF,

⇒ BC = EF [Given]

⇒ AC = DF [Given]

⇒ ∠C = ∠F [Given]

∴ △ABC ≅ △DEF (By S.A.S axiom)

Hence, △ABC and △DEF are congruent by S.A.S axiom.

(b) Given,

In △ABC and △PQR,

⇒ AB = PQ [Given]

⇒ BC = QR [Given]

⇒ ∠C = ∠R [Given]

Here the equal angles are not the included angles, thus the triangles are not necessarily congruent.

Hence, △ABC and △PQR are not necessarily congruent.

(c) Given,

In △ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 90° + ∠B + 40° = 180°

⇒ ∠B + 130° = 180°

⇒ ∠B = 180° - 130°

⇒ ∠B = 50°.

In △ABC and △LMN,

⇒ AB = LM [Given]

⇒ ∠B = ∠M [Both equal to 50°]

⇒ ∠A = ∠L [Both equal to 90°]

∴ △ABC ≅ △LMN (By A.S.A axiom)

Hence, △ABC and △LMN are congruent by A.S.A axiom.

(d) Given,

In △ABC and △DEF,

⇒ ∠B = ∠E = 90°

⇒ AC = DF

⇒ AB = DE

∴ △ABC ≅ △DEF (By R.H.S. axiom)

Hence, △ABC and △DEF are congruent by R.H.S. axiom.

In the given figure, P is a point in the interior of ∠ABC. If PL ⊥ BA and PM ⊥ BC such that PL = PM, prove that BP is the bisector of ∠ABC.

Answer

Given,

In △BLP and △BMP,

⇒ ∠L = ∠M = 90°

⇒ PL = PM [Given]

⇒ PB = PB [Common side]

∴ △BLP ≅ △BMP (By R.H.S. axiom)

⇒ ∠LBP = ∠PBM [Corresponding angles of congruent triangles are equal.]

Hence, proved that BP is the bisector of ∠ABC.

In the given figure, equal sides BA and CA of △ABC are produced to Q and P respectively such that AP = AQ. Prove that PB = QC.

Answer

In △ ABC,

⇒ BA = CA [Given]

⇒ ∠ABC = ∠ACB [Angles opposite to equal sides are equal]

From figure,

⇒ ∠PAB + ∠CAB = 180° [Linear pair of angles]

⇒ ∠PAB = 180° - ∠CAB ......(1)

⇒ ∠QAC + ∠CAB = 180° [Linear pair of angles]

⇒ ∠QAC = 180° - ∠CAB ......(2)

From eq.(1) and (2), we have:

⇒ ∠PAB = ∠QAC

In △PAB and △QAC,

⇒ ∠PAB = ∠QAC [Proved above]

⇒ AP = AQ [Given]

⇒ BA = CA [Given]

∴ △PAB ≅ △QAC (By S.A.S axiom)

∴ PB = QC [Corresponding angles of congruent triangles are equal]

Hence, proved that PB = QC.

In the given figure, median AD of △ABC is produced. If BL and CM are perpendiculars drawn on AD and AD produced, prove that BL = CM.

Answer

In △ABC,

AD is the median on side BC

⇒ BD = DC

In △LDB and △CDM,

⇒ BD = DC [Given, AD is the median]

⇒ ∠L = ∠M [Each equal to 90°]

⇒ ∠LDB = ∠CDM [Vertically opposite angles are equal]

∴ △LDB ≅ △CDM (By A.A.S. axiom)

⇒ BL = CM [Corresponding part of congruent triangles are equal.]

Hence, proved that BL = CM.

In the given figure, M is the mid-point of AB and CD. Prove that CA = BD and CA || BD.

Answer

Given,

M is the mid-point of AB and CD.

Thus, MC = MD and MB = MA

In △MAC and △MBD,

⇒ MA = MB [Proved above]

⇒ MC = MD [Proved above]

⇒ ∠CMA = ∠BMD [Vertically opposite angles are equal]

∴ △MAC ≅ △MBD (By S.A.S. axiom)

⇒ CA = BD [Corresponding part of congruent triangles are equal.]

⇒ ∠CAM = ∠DBM [Corresponding part of congruent triangles are equal.]

Since, ∠CAM and ∠DBM are alternate angles and since they are equal,

∴ CA || BD

Hence, proved that CA = BD and CA || BD.

In the given figure, PA ⊥ AB; QB ⊥ AB and PA = QB. If PQ intersects AB at M, show that M is the mid-point of both AB and PQ.

Answer

Given,

⇒ PA = QB

From figure,

⇒ ∠PAM = ∠QBM = 90°

In △MAP and △MBQ,

⇒ ∠PAM = ∠QBM [Both equal to 90°]

⇒ AP = BQ [Given]

⇒ ∠AMP = ∠QMB [Vertically opposite angles are equal]

∴ △MAP ≅ △MBQ (By A.A.S axiom)

⇒ AM = MB [Corresponding parts of congruent triangles are equal]

⇒ PM = MQ [Corresponding parts of congruent triangles are equal]

Thus, M is the mid-point of both AB and PQ.

Hence, proved that M is the mid-point of both AB and PQ.

AB is a line segment. AX and BY are two equal line segments drawn on opposite sides of AB such that AX || YB. If AB and XY intersect at M, prove that :

(i) △AMX ≅ △BMY

(ii) AB and XY bisect each other at M.

Answer

(i) In △AMX and △BMY,

⇒ ∠AXM = ∠BYM [Alternate angles are equal as AX || YB]

⇒ AX = BY [Given]

⇒ ∠AMX = ∠BMY [Vertically opposite angles are equal]

∴ △AMX ≅ △BMY (By A.A.S axiom)

Hence, proved that △AMX ≅ △BMY.

(ii) As, △AMX ≅ △BMY

⇒ AM = MB [Corresponding parts of congruent triangles are equal]

∴ M is the mid-point of line segment AB.

⇒ XM = MY [Corresponding parts of congruent triangles are equal]

∴ M is the mid-point of line segment XY.

Hence, proved that AB and XY bisect each other at M.

In the given figure, the sides BA and CA of △ABC have been produced to D and E such that BA = AD and CA = AE. Prove that, ED || BC.

Answer

In △ABC and △ADE,

⇒ AB = AD [Given]

⇒ AC = AE [Given]

⇒ ∠BAC = ∠EAD [Vertically opposite angles are equal]

∴ △ABC ≅ △ADE [By S.A.S axiom]

⇒ ∠ABC = ∠ADE [Corresponding parts of congruent triangles are equal.]

From figure,

∠ABC and ∠ADE are alternate angles and since they are equal.

∴ ED || BC

Hence, proved that ED || BC.

In the given figure, the line segments AB and CD intersect at a point M in such a way that AM = MD and CM = MB. Prove that, AC = BD but AC may not be parallel to BD.

Answer

In △MAC and △MDB,

⇒ AM = MD [Given]

⇒ CM = MB [Given]

⇒ ∠AMC = ∠DMB [Vertically opposite angles are equal]

∴ △MAC ≅ △MDB (By S.A.S axiom)

⇒ AC = BD [Corresponding parts of congruent triangles are equal.]

⇒ ∠MDB = ∠MAC [Corresponding parts of congruent triangles are equal.]

∴ ∠MDB ≠ ∠MCA

Thus, we cannot prove that AC // BD.

Hence, proved that AC = BD but AC may not be parallel to BD.

If two altitudes of a triangle are equal, prove that it is an isosceles triangle.

Answer

Given,

CE = BD [Altitudes of △ABC are equal]

In △ABD and △ACE,

⇒ ∠ADB = ∠AEC [Each equal to 90°]

⇒ CE = BD [Given]

⇒ ∠DAB = ∠EAC [Common angle]

∴ △ABD ≅ △ACE (∵ A.A.S. axiom)

⇒ AB = AC [Corresponding parts of congruent triangles are equal]

Since, two sides are equal.

Hence, △ABC is an isosceles triangle with AB = AC.

In the given figure, ∠BAC = ∠CDB and ∠BCA = ∠CBD. Prove that AB = CD.

Answer

In △DCB and △ABC,

⇒ ∠DBC = ∠ACB [Given]

⇒ ∠BDC = ∠BAC [Given]

⇒ BC = BC [Common side]

∴ △ABC ≅ △DCB (By A.A.S. axiom)

⇒ AB = DC [Corresponding parts of congruent triangles are equal]

Hence, proved that AB = CD.

In the given figure, ∠ABD = ∠EBC, BD = BC and ∠ACB = ∠EDB. Prove that AB = BE.

Answer

Given,

∠ACB = ∠EDB and ∠ABD = ∠EBC

From figure,

⇒ ∠EBD = ∠ABE + ∠ABD .....(1)

Also,

⇒ ∠ABC = ∠ABE + ∠EBC

⇒ ∠ABC = ∠ABE + ∠ABD .....(2)

From eq.(1) and (2), we have :

⇒ ∠ABC = ∠EBD

In △EDB and △ACB,

⇒ DB = BC [Given]

⇒ ∠ABC = ∠EBD [Proved above]

⇒ ∠ACB = ∠EDB [Given]

∴ △EDB ≅ △ACB (By A.S.A. axiom)

⇒ AB = BE [Corresponding parts of congruent triangles are equal.]

Hence, proved that AB = BE.

In the given figure, AY ⊥ ZY nd BY ⊥ XY such that AY = ZY and BY = XY. Prove that AB = ZX.

Answer

⇒ ∠AYZ = ∠XYB [Each equal to 90°]

Adding ∠AYX on both L.H.S and R.H.S, we have:

⇒ ∠AYZ + ∠AYX = ∠XYB + ∠AYX

⇒ ∠XYZ = ∠AYB

In △XYZ and △AYB,

⇒ ZY = AY [Given]

⇒ XY = BY [From figure]

⇒ ∠XYZ = ∠AYB [Proved above]

∴ △XYZ ≅ △AYB (By S.A.S axiom)

⇒ AB = ZX [Corresponding parts of congruent triangles are equal]

Hence, proved that AB = ZX.

In the given figure, ABCD is a square and △PAB is an equilateral triangle.

(i) Prove that △APD ≅ △BPC.

(ii) Show that ∠DPC = 15°.

Answer

(i) We know that,

Each interior angle in a square is 90° and each interior angle in an equilateral triangle is 60°.

From figure,

⇒ ∠DAP = ∠DAB + ∠BAP

⇒ ∠DAP = 90° + 60° = 150°

⇒ ∠CBP = ∠CBA + ∠ABP

⇒ ∠CBP = 90° + 60° = 150°

In △APD and △BPC,

⇒ ∠DAP = ∠CBP [Each equal to 150°]

⇒ AD = BC [Sides of a square]

⇒ AP = BP [Sides of an equilateral triangle]

∴ △APD ≅ △BPC (By S.A.S axiom)

Hence, △APD ≅ △BPC.

(ii) ABCD is a square.

∴ AB = AD = DC = BC

APB is an equilateral triangle.

∴ AP = PB = AB

So, we get :

AP = AD and PB = BC

∴ △APD and △BPC are isosceles triangle.

We know that,

Angles opposite to equal sides are equal.

∴ ∠APD = ∠ADP = x (let) and ∠BPC = ∠BCP = y (let)

In △APD,

By angle sum property of triangle,

⇒ ∠APD + ∠ADP + ∠DAP = 180°

⇒ x + x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$

⇒ x = 15°.

In △BPC,

By angle sum property of triangle,

⇒ ∠BPC + ∠BCP + ∠PBC = 180°

⇒ y + y + 150° = 180°

⇒ 2y = 180° - 150°

⇒ 2y = 30°

⇒ y = $\dfrac{30°}{2}$

⇒ y = 15°

From figure,

⇒ ∠PDC = ∠ADC - ∠ADP = 90° - 15° = 75°

⇒ ∠PCD = ∠BCD - ∠BCP = 90° - 15° = 75°

In △DPC,

By angle sum property of triangle,

⇒ ∠DPC + ∠PDC + ∠PCD = 180°

⇒ ∠DPC + 75° + 75° = 180°

⇒ ∠DPC + 150° = 180°

⇒ ∠DPC = 180° - 150°

⇒ ∠DPC = 30°.

Hence, proved that ∠DPC = 30°.

In the given figure, in △ABC, ∠B = 90°. If ABPQ and ACRS are squares, prove that:

(i) △ACQ ≅ △ABS

(ii) CQ = BS.

Answer

(i) From figure,

In △ACQ,

⇒ ∠BAC = ∠QAC - ∠QAB

⇒ ∠QAC = ∠BAC + ∠QAB

⇒ ∠QAC = ∠BAC + 90° ....(1)

In △ABS,

⇒ ∠BAC = ∠BAS - ∠CAS

⇒ ∠BAS = ∠BAC + ∠CAS

⇒ ∠BAS = ∠BAC + 90° ....(2)

From eq.(1) and (2), we have:

⇒ ∠QAC = ∠BAS

In △ACQ and △ABS,

⇒ ∠QAC = ∠BAS [Proved above]

⇒ AQ = AB [Sides of square ABPQ]

⇒ AC = AS [Sides of square ACRS]

∴ △ACQ ≅ △ABS (By S.A.S axiom)

Hence, proved that △ACQ ≅ △ABS.

(ii) As,

△ACQ ≅ △ABS

∴ CQ = BS [Corresponding parts of congruent triangles are equal]

Hence, proved that CQ = BS.

Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:

(i) ∠SAQ = ∠ABC

(ii) SQ = AC.

Answer

(i) Given,

ABPQ and ADRS are square.

Each angle of a square = 90°

From figure,

⇒ ∠SAQ + ∠SAD + ∠BAD + ∠BAQ = 360°

⇒ ∠SAQ + 90° + ∠BAD + 90° = 360°

⇒ ∠SAQ + ∠BAD + 180° = 360°

⇒ ∠SAQ + ∠BAD = 360° - 180°

⇒ ∠SAQ = 180° - ∠BAD ...(1)

In parallelogram ABCD,

⇒ ∠ABC + ∠BAD = 180° (Sum of adjacent angles of a // gm = 180°)

⇒ ∠ABC = 180° - ∠BAD ....(2)

From eq.(1) and (2), we have:

⇒ ∠SAQ = ∠ABC

Hence, proved that ∠SAQ = ∠ABC.

(ii) In square ADRS,

AS = SR = RD = AD

⇒ AD = BC (Opposite sides of a parallelogram ABCD are equal)

∴ AS = BC

In △SAQ and △CBA,

⇒ ∠SAQ = ∠ABC (Proved above)

⇒ AS = BC (Proved above)

⇒ AQ = AB (Sides of a square ABPQ)

∴ △SAQ ≅ △CBA (By S.A.S axiom)

⇒ SQ = AC (Corresponding parts of congruent triangles are equal)

Hence, proved that SQ = AC.

In the given figure, ABCD is a parallelogram, E is the mid-point of BC. DE produced meets AB produced at L. Prove that:

(i) AB = BL

(ii) AL = 2DC

Answer

(i) Given,

ABCD is a parallelogram.

We know that,

Opposite sides of a parallelogram are equal.

⇒ AB = CD and AD = BC

In △DEC and △BEL,

⇒ ∠LBE = ∠DCE (Alternate angles, since AB || DC)

⇒ EC = EB (Given)

⇒ ∠DEC = ∠BEL (Vertically opposite angles are equal)

∴ △DEC ≅ △BEL (By A.S.A axiom)

⇒ DC = BL (Corresponding parts of congruent triangles are equal)

Since, AB = DC

∴ AB = BL.

Hence, proved that AB = BL.

(ii) From figure,

⇒ AL = AB + BL

⇒ AL = DC + DC

⇒ AL = 2DC.

Hence, proved that AL = 2DC.

Equilateral triangle ABD and ACE are drawn on the sides AB and AC of △ABC as shown in the figure. Prove that :

(i) ∠DAC = ∠EAB

(ii) DC = BE

Answer

(i) Given,

△ABD and △ACE are equilateral triangles.

⇒ ∠DAB = ∠ABD = ∠BDA = ∠EAC = ∠ACE = ∠CEA = 60°

From figure,

∠DAC = ∠DAB + ∠BAC

⇒ ∠DAC = 60° + ∠BAC .....(1)

∠EAB = ∠EAC + ∠BAC

⇒ ∠EAB = 60° + ∠BAC .....(2)

From eq.(1) and (2), we have :

⇒ ∠EAB = ∠DAC

Hence, proved that, ∠EAB = ∠DAC.

(ii) In △DAC and △BAE,

⇒ ∠EAB = ∠DAC (Proved above)

⇒ AD = AB (Sides of an equilateral triangle)

⇒ AC = AE (Sides of an equilateral triangle)

∴ △DAC ≅ △BAE (By S.A.S axiom)

∴ DC = BE (Corresponding parts of congruent triangles are equal)

Hence, proved that DC = BE.

In the given figure, ABCD is a square and P, Q, R are points on AB, BC and CD respectively such that AP = BQ = CR and ∠PQR = 90°. Prove that:

(i) PB = QC

(ii) PQ = QR

(iii) ∠QPR = 45°

Answer

(i) Given,

ABCD is a square.

AB = BC = CD = DA

Given,

AP = BQ = CR

From figure,

⇒ PB = AB - AP ....(1)

⇒ QC = BC - BQ

⇒ QC = AB - AP [As, BC = AB and BQ = AP] ....(2)

From eq.(1) and (2), we have :

⇒ PB = QC

Hence, proved that PB = QC.

(ii) In △PBQ and △QCR,

⇒ ∠PBQ = ∠QCR (Both equal to 90°)

⇒ BQ = CR (Given)

⇒ PB = QC (Proved above)

∴ △PBQ ≅ △QCR (By S.A.S axiom)

∴ PQ = QR (Corresponding parts of congruent triangles are equal)

Hence, proved that PQ = QR.

(iii) In △QPR,

⇒ PQ = QR

∴ △PQR is an isosceles triangle.

⇒ ∠QPR = ∠QRP = f (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠QPR + ∠QRP + ∠PQR = 180°

⇒ f + f + 90° = 180°

⇒ 2f = 180° - 90°

⇒ 2f = 90°

⇒ f = $\dfrac{90°}{2}$

⇒ f = 45°.

⇒ ∠QPR = ∠QRP = 45°.

Hence, proved that ∠QPR = 45°.

In the given figure, ABCD is a square, EF || BD and R is the mid-point of EF. Prove that:

(i) BE = DF

(ii) AR bisects ∠BAD

(iii) If AR is produced, it will pass through C.

Answer

(i) Given,

EF || BD

DC is the transversal.

⇒ ∠BDC = ∠EFC

BC is the transversal.

⇒ ∠DBC = ∠FEC

Given,

ABCD is a square.

⇒ ∠A = ∠B = ∠C = ∠D = 90°

In △BDC,

BC = DC [Sides of a square are equal]

⇒ ∠DBC = ∠CDB = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠DBC + ∠BCD + ∠CDB = 180°

⇒ x + 90° + x = 180°

⇒ 2x = 180° - 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°

⇒ ∠DBC = ∠CDB = 45°

∴ ∠DBC = ∠CDB = ∠FEC = ∠EFC = 45°.

⇒ △BDC and △EFC are isosceles triangles.

⇒ EC = CF (As, ∠FEC = ∠EFC)

From figure,

⇒ BE = BC - EC .....(1)

⇒ DF = DC - CF

As, BC = DC and EC = CF

⇒ DF = BC - EC .....(2)

From (1) and (2), we get :

⇒ BE = DF.

Hence, proved that BE = DF.

(ii) In △ABE and △ADF,

⇒ ∠ABE = ∠ADF (Both equal to 90°)

⇒ AB = AD (Sides of a square)

⇒ BE = DF (Proved above)

∴ △ABE ≅ △ADF (By S.A.S axiom)

⇒ ∠BAE = ∠DAF (Corresponding parts of congruent triangles are equal)

⇒ AE = AF (Corresponding parts of congruent triangles are equal)

∴ △AEF is an isosceles triangle.

Given,

R is the mid-point of EF.

We know that,

In an isosceles triangle, the median to the base is also the angle bisector of the vertex angle.

∴ AR bisects ∠EAF.

⇒ ∠EAR = ∠FAR

From figure,

⇒ ∠BAR = ∠BAE + ∠EAR ....(1)

⇒ ∠DAR = ∠DAF + ∠FAR

Since,

∠EAR = ∠FAR and ∠BAE = ∠DAF.

⇒ ∠DAR = ∠BAE + ∠EAR .....(2)

From eq.(1) and (2), we have:

⇒ ∠BAR = ∠DAR.

∴ AR bisects ∠BAD.

Hence, proved that AR bisects ∠BAD.

(iii) Since, AR bisects ∠BAD

AR lies on diagonal AC, as diagonals of a square bisect the vertex angle.

∴ AR if produced, must pass through the vertex C.

Hence, proved that, if AR is produced, it will pass through C.

ABCD is a parallelogram in which ∠A and ∠C are obtuse. Points X and Y are taken on diagonal BD such that ∠AXD = ∠CYB = 90°. Prove that : XA = YC.

Answer

AD || BC and BD is the transversal.

⇒ ∠ADX = ∠CBY (Alternate interior angles are equal)

In △XAD and △YCB,

⇒ ∠ADX = ∠CBY (Proved above)

⇒ ∠AXD = ∠BYC (Both equal to 90°)

⇒ AD = BC (Opposite sides of a parallelogram are equal)

∴ △XAD ≅ △YCB (By A.A.S. axiom)

⇒ XA = YC (Corresponding parts of congruent triangles are equal)

Hence, proved 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 △BEC ≅ △DCF.

Answer Given,

AD = DF ....(1)

AB = BE ....(2)

We know that,

Opposite sides of parallelogram are equal.

∴ AD = BC ....(3)

∴ AB = CD ....(4)

From eq.(1) and (3), we get :

⇒ BC = DF

From eq.(2) and (4), we get :

⇒ BE = CD

We know that,

Opposite angles of a parallelogram are equal.

⇒ ∠ABC = ∠ADC = x (let)

From figure,

Since, AE is a straight line.

⇒ ∠CBE + ∠ABC = 180°

⇒ ∠CBE + x = 180°

⇒ ∠CBE = 180° - x .......(5)

Since, AF is a straight line.

⇒ ∠CDF + ∠ADC = 180°

⇒ ∠CDF + x = 180°

⇒ ∠CDF = 180° - x ........(6)

From eq.(5) and (6), we get :

⇒ ∠CBE = ∠CDF

In △BEC and △DCF,

⇒ BE = CD (Proved above)

⇒ ∠CBE = ∠CDF (Proved above)

⇒ BC = DF (Proved above)

∴ △BEC ≅ △DCF (By S.A.S axiom)

Hence, proved that △BEC ≅ △DCF.

The perpendicular bisectors of the sides of a △ABC meet at I. Prove that : IA = IB = IC.

Answer From figure,

AL, CN and BM are perpendicular bisectors of sides BC, AB and AC respectively.

In △BIL and △CIL,

⇒ BL = LC (Since, AL is the perpendicular bisector of BC)

⇒ ∠BLI = ∠CLI (Both equal to 90°)

⇒ LI = LI (Common side)

∴ △BIL ≅ △CIL (By S.A.S axiom)

⇒ IB = IC .....(1) (Corresponding parts of congruent triangles are equal)

In △CIM and △AIM,

⇒ CM = AM (Since, BM is the perpendicular bisector of AC)

⇒ ∠CMI = ∠AMI (Both equal to 90°)

⇒ MI = MI (Common side)

∴ △CIM ≅ △AIM (By S.A.S axiom)

⇒ IC = IA .....(2) (Corresponding parts of congruent triangles are equal)

From eq.(1) and (2), we get:

⇒ IA = IB = IC.

Hence, proved that IA = IB = IC.