On the sides AB and AC of triangle ABC, equilateral triangle ABD and ACE are drawn.
Prove that: (i) ∠CAD = ∠BAE
(ii) CD = BE
Given: ΔABD is an equilateral triangle.
ΔACE is an equilateral triangle
We need to prove that
(i) ∠CAD = ∠BAE
Proof:
(i) ΔABD is equilateral
∴ Each angel = 60°
⇒ ∠BAD = 60° ...(1)
Similarly,
ΔACE is equilateral
∴ Each angel = 60°
⇒ ∠CAE = 60° ...(2)
⇒ ∠BAD = ∠CAE ...[ from (1) and (2) ]...(3)
Adding ∠BAC to both sides, we have
⇒ ∠BAD + ∠BAC = ∠CAE + ∠BAC
⇒ ∠CAD = ∠BAE ...(4)
(ii) In ΔCAD and ΔBAE
AC = AE ...[ ΔACE is equilateral ]
∠CAD = ∠BAE ... [ from (4) ]
AD = AB ...[ ΔABD is euilateral ]
∴ By Side-Angel-Side criterion of congruency,
ΔCAD ≅ ΔBAE
The corresponding parts of the congruent
triangles are congruent.
∴ CD = BE ...[ by c.p.c.t ]
Hence proved.
In the following diagram, ABCD is a square and APB is an equilateral triangle.
(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC.
Given: ABCD is a Square and ΔAPB is an equilateral triangle.
We need to
(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC
(i) Proof:
AP = PB = AB ...[ APB is an equilateral triangle ]
Also, we have,
∠PBA = ∠PAB = ∠APB = 60° ...(1)
Since ABCD is a square, we have
∠A =∠ B = ∠C = ∠D = 90° ...(2)
Since ∠DAP = ∠A - PAB ...(3)
⇒ ∠DAP = 90° - 60°
⇒ ∠DAP =30° ...[ from (1) and (2) ] ...(4)
Similarly ∠CBP = ∠B - ∠PBA
⇒ ∠CBP = 90° - 60°
⇒ ∠CBP = 30° ...[ from (1) and (2) ] ...(5)
⇒ ∠DAP = ∠CBP ....[ from (1) and (2) ] ...(6)
In ΔAPD and ΔBPC
AD = BC ...[ Sides of square ABCD ]
∠DAP = ∠CBP ...[ from(6) ]
AP= BP [ Sides of equilateral ΔAPB ]
∴ By Side-Angel-Side Criterion of Congruence, we have,
ΔAPD ≅ ΔBPC
(ii)
AP = PB = AB ....[ ΔAPB is an equilateral triangle ] ...(7)
AB = BC = CD = DA ...[ Sides of square ABCD ] ...(8)
From (7) and (8), we have
AP = DA aand PB = BC ... (9)
In ΔAPD,
AP = DA ...[ from (9) ]
∠ADP = ∠APD ...[ Angel opposite to equal sides are equal ] ...(10)
∠ADP + ∠APD+ +∠DAP + 180° ...[ Sum of angel of a triangle = 180° ]
⇒ ∠ADP + ∠ADP + 30° = 180° [ from (3), ∠DAP =30° from (10), ∠ADP = ∠APD ]
⇒ ∠ADP + ∠ADP = 180° - 30°
⇒ 2∠ADP =
⇒∠ADP= 75°
We have ∠PDC =∠D - ∠ADP
⇒∠PDC = 90° - 75°
⇒∠PDC =15° ...(11)
In BPC,
PB = BC ...[ from (9) ]
∴ ∠PCB =∠ BPC ...[Angel opposite to equal sides are equal ] ... (12)
∠PCB + ∠BPC + ∠CBP = 180° ....[ Sum of angel of a triangle = 180° ]
⇒ ∠PCB + ∠PCB + 30° = 180° ....[ from (5), ∠CBP = from (12) , ∠PCB =∠BPC ]
⇒ 2∠PCB = 180° - 30°
⇒ 2∠PCB =
⇒ ∠PCB = 75°
We have ∠PCD = ∠C - ∠PCB
⇒ ∠PCD = 90° - 75°
⇒ ∠PCD = 15° ... (13)
In ΔDPC,
∠PDC = 15°
∠PCD = 15°
∠PCD + ∠PDC + ∠DPC = 180° ...[ Sum of angles of a triangle = 180° ]
⇒ 15° + 15° + ∠DPC = 180°
⇒ ∠DPC = 180° - 30°
⇒ ∠DPC = 150°
∴ Angles of DPC, are: 15° , 150° , 15°
In the following diagram, ABCD is a square and APB is an equilateral triangle.
(i) Prove that: ΔAPD ≅ ΔBPC
(ii) Find the angles of ΔDPC.
Given: ABCD is a Square and ΔAPB is an equilateral triangle.
(i) Proof: In ΔAPB,
AP = PB = AB ...[ APB is an equilateral triangle ]
Also, we have,
∠PBA = ∠PAB = ∠APB = 60° ...(1)
Since ABCD is a square, we have
∠A =∠ B = ∠C = ∠D = 90° ...(2)
Since ∠DAP = ∠A + ∠PAB ..(3)
⇒ ∠DAP = 90° + 60°
⇒ ∠DAP = 150° ...[ from (1) and (2) ] ...(4)
Similarly ∠CBP = ∠B + ∠PBA
⇒ ∠CBP = 90° + 60°
⇒ ∠CBP = 150° ...[ from (1) and (2) ] ...(5)
⇒ ∠DAP = ∠CBP ....[ from (1) and (2) ] ...(6)
In ΔAPD and ΔBPC
AD = BC ...[ Sides of square ABCD ]
∠DAP = ∠CBP ...[ from(6) ]
AP= BP [ Sides of equilateral ΔAPB ]
∴ By Side-AAngel-SIde Criterion of Congruence, we have,
ΔAPD ≅ ΔBPC
(ii)
AP = PB = AB ....[ ΔAPB is an equilateral triangle ] ...(7)
AB = BC = CD = DA ...[ Sides of square ABCD ] ...(8)
From (7) and (8), we have
AP = DA aand PB = BC ... (9)
In ΔAPD,
AP = DA ...[ from (9) ]
∠ADP = ∠APD ...[ Angel opposite to equal sides are equal ] ...(10)
∠ADP + ∠APD+ +∠DAP + 180° ...[ Sum of angel of a triangle = 180° ]
⇒ ∠ADP + ∠ADP + 150° = 180° [ from (3), ∠DAP =150° from (10), ∠ADP = ∠APD ]
⇒ ∠ADP + ∠ADP = 180° - 150°
⇒ 2∠ADP = 30°
⇒ ∠ADP =
⇒∠ADP= 15°
We have ∠PDC =∠D - ∠ADP
⇒∠PDC =90° - 15°
⇒∠PDC =75° ...(11)
In ΔBPC,
PB = BC ...[ from (9) ]
∴ ∠PCB =∠ BPC ...[Angel opposite to equal sides are equal ] ... (12)
∠PCB + ∠BPC + ∠CBP = 180° ....[ Sum of angel of a triangle = 180° ]
⇒ ∠PCB + ∠PCB + 30° = 180° ....[ from (5), ∠CBP = 150° from (12) , ∠PCB =∠BPC ]
⇒ 2∠PCB = 180° - 150°
⇒ 2∠PCB =
⇒ ∠PCB = 15°
We have ∠PCD = ∠C - ∠PCB
⇒ ∠PCD = 90° - 15°
⇒ ∠PCD = 75° ... (13)
In ΔDPC,
∠PDC = 75°
∠PCD = 75°
∠PCD + ∠PDC + ∠DPC = 180° ...[ Sum of angles of a triangle = 180° ]
⇒ 75° + 75° + ∠DPC = 180°
⇒ ∠DPC = 180° - 150°
⇒ ∠DPC = 30°
∴ Angles of DPC, are: 75°, 30° , 75°
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.
Given: A(Δ ABC) is right-angled at B.
ABPQ and ACRS are squares
To Prove:
(i) ΔACQ ≅ ΔASB
(ii) CQ = BS
Proof:
(i)
∠ QAB = 90° ...[ ABPQ is a square ] ...(1)
∠ SAC = 90° ...[ ACRS is a square ] ...(2)
From (1) and (2) , We have
∠ QAB = ∠SAC ...(3)
Adding ∠BAC to both sides of (3), We have
∠ QAB + ∠BAC = ∠SAC + ∠BAC
⇒ ∠QAC = ∠SAB ...(4)
In ΔACQ and ΔASB,
QA = QB ...[ Sides of a square ABPQ ]
∠QAC = ∠SAB ...[ From(4) ]
AC = AS ...[ sides of a square ACRS ]
∴ By Side -Angle-Side criterion of congruence,
ΔACQ ≅ ΔASB
(ii)
The corresponding parts of the congruent triangles are congruent,
∴ CQ = BS ...[ c.p.c.t. ]
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.
Given: A(ΔABC) in which BD is the median to AC.
BD is produced to E such that BD = DE,
We need to prove that AE II BC.
Construction: Join AE
Proof:
AD = DC ...[ BD is median to AC ] ...(1)
In ΔBDC and ΔADE,
BD = DE ...[ Given ]
∠BDC = ∠ADE = 90° ...[ Vertically opposite angles ]
AD = DC ...[ from(1) ]
∴ By Side-Angle-Side Criterion of congruence,
ΔBDC ≅ ΔADE
The corresponding parts of the congruent triangles are congruent.
∴ ∠EAD = ∠BCD ...[ c.p.c.t. ]
But these are alternate angles and AC is the transversal.
Thus, AE || 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: (i) ΔXTQ ≅ ΔXSQ.
(ii) PX bisects angle P.
Given: A( ΔPQR ) in which QX is the bisector of ∠Q. and RX is the bisector of ∠R.
XS ⊥ QR and XT ⊥ PQ.
We need to prove that
(i) ΔXTQ ≅ ΔXSQ.
(ii) PX bisects angle P.
Construction: Draw XZ ⊥ PR and join PX.
Proof:
(i) In ΔXTQ and ΔXSQ,
∠QTX = ∠QSX = 90° ...[ XS ⊥ QR and XT ⊥ PQ ]
∠TQX = ∠SQX ...[ QX is bisector of ∠Q ]
QX = QX ...[ Common ]
∴ By Angle-Angle-Side Criterion of congruence,
ΔXTQ ≅ ΔXSQ
(ii) The corresponding parts of the congruent triangles are congruent.
∴ XT = XS ...[ c.p.c.t. ]
In ΔXSR ≅ ΔXZR
∠XSR = ∠XZR = 90° ...[ XS ⊥ QR and ∠XSR = 90° ]
∠SRX = ∠ZRX ...[ RX is bisector of ∠R ]
RX = RX ....[ Common ]
∴ By Angle-Angle-Side criterion of congruence,
ΔXSR ≅ ΔXZR
The corresponding parts of the congruent triangles are congruent.
∴ XS = XZ ...[ c.p.c.t. ] ...(2)
From (1) and (2)
XT = XZ ....(3)
In ΔXTP and ΔXZP
∠XTP = ∠XZP = 90° ....[ Given ]
Hyp. XP = Hyp. XP ....[ Common ]
XT = XZ ....[ from(3) ]
∴ By Right angle-Hypotenuse-side criterion of congruence,
ΔXTP ≅ ΔXZP
The corresponding parts of the congruent triangles are congruent.
∴ ∠XPT = ∠XPZ ...[ c.p.c.t. ]
∴ PX bisects ∠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 in which ∠A and ∠C are obtuse.
Points X and Y are taken on the diagonal BD.
Such that ∠XAD = ∠YCB = 90°.
We need to prove that XA = YC
Proof:
ln ΔXAD and ΔYCB
∠XAD = ∠YCB= 90° ...[ Given ]
AD = BC ...[ Opposite sides of a parallelogram ]
∠ADX = ∠CBY ...[ Alternate angles ]
∴ By Angle-Side-Angle criterion of congruence,
ΔXAD ≅ ΔYCB
The corresponding parts of the congruent triangles are congruent.
∴ XA = YC ...[ c.p.c.t. ]
Hence proved.
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: ΔBEC ≅ ΔDCF.
ABCD is a parallelogram, The sides AB and AD are produced to E and F respectively,
such that AB = BE and AD = DF
We need to prove that ΔBEC ≅ ΔDCF.
Proof:
AB = DC ...[ Opposite sides of a parallelogram ] ...(1)
AB = BE ...[ Given ] ...(2)
From (1) and (2), We have
BF = DC ...(3)
AD = BC ...[ Opposite sides of a parallelogram ] ...(4)
AD = DF ....[Given]
From (4) and (5), we have
BC = DF ...(6)
Since AD II BC, the corresponding angles are equal.
∴ ∠DAB = ∠CBE ...(7)
Since AB II DC, the corresponding angles are equal.
∴ ∠DAB = ∠FDC ...(8)
From (7) and (8), we have
∠CBE = ∠FDC
ln ΔBEC and ΔDCF
BF = DC ....[ from (3) ]
∠CBE = ∠FDC ...[ from (9) ]
BC = DF ....[ from (6) ]
∴ By Side-Angle-Side criterion of congruence,
ΔBEC ≅ ΔDCF
Hence proved.
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 ΔABC and ΔPQR are congruent.
Since, BC = QR, We have
BD = QS and DC = SR ....[ D is the mid-point of BC and S is the mid-point of QR ]
In ΔABD and ΔPQS,
AB = PQ ...(1)
AD = PS ...(2)
BD = QS ...(3)
Thus, by Side-Side-Side criterion of congruence,
We have ΔABD ≅ ΔPQS
Similarly, in ΔADC and ΔPSR
AD = PS ...(4)
AC = PR ...(5)
DC = SR ....(6)
Thus, by Side-Side-Side criterion of congruence,
We have ΔADC ≅ ΔPSR
We have
BC = BD + DC ...[ D is the mid-point of BC ]
= QS + SR ...[ From (3) and (6) ]
= QR ....[ S is the mid-point of QR ] ...(7)
Now consider the triangles ΔABC and ΔPQR
AB = PQ ...[ from(1) ]
BC = QR ...[ from(7) ]
AC = PR ...[ from(7) ]
∴ By Side-Side-Side criterion of congruence, we
have ΔABC ≅ ΔPQR
Hence proved.
In the following diagram, AP and BQ are equal and parallel to each other.
Prove that:
(i) ΔAOP≅ ΔBOQ.
(ii) AB and PQ bisect each other.
In the figure, AP and BQ are equal and parallel to each other.
∴ AP = BQ and AP || BQ.
We need to prove that
(i) ΔAOP≅ ΔBOQ.
(ii) AB and PQ bisect each other
(i) ∵ AP || BQ
∴∠APO =∠BOQ ...[ Alternate angles ] ...(1)
and ∠PAO =∠QBO ...[ Alternate angles ] ...(2)
Now in ΔAOP and ΔBOQ.
∠APO =∠BQO ...[ from (1) ]
AP = BQ ...[ given ]
∠PAO = ∠QBO ...[ from (1) ]
∴ By Angel-Side-Angel criterion of congruence, we have
ΔAOP≅ ΔBOQ.
(ii) The corresponding parts of the congruent triangles are congruent.
∴ OP = OQ ...[ c. p. c .t ]
OA = OB ...[ c. p. c .t ]
Hence AB and PQ bisect each other.
In the following figure, OA = OC and AB = BC.
Prove that:
(i) ∠AOB = 90o
(ii) ΔAOD ≅ ΔCOD
(iii) AD = CD
Given:
In the figure, OA=OC, AB =BC
We need to prove that,
AOB = 90°
(i) In ΔABO and ΔCBO,
AB = BC ...[given ]
AO = CO ...[ given ]
OB = OB ...[ common ]
∴By Side-Side-Side criterion of congruence, we have
ΔABO ≅ ΔCBO
The corresponding parts of the congruent triangles are congruent.
∴∠ABO = ∠CBO ...[c. p.c.t. ]
⇒ ∠ABD = ∠CBD
and ∠AOB = ∠COB ...[c. p.c t ]
We have
∠AOB + ∠COB = 180° .....[ linear pair ]
⇒ ∠AOB = ∠ COB= 90° and AC ⊥ BD
(ii) In ΔAOD and ΔCOD,
OD = OD ...[ common ]
∠AOD = ∠COD ...[ each=90° ]
AO = CO ...[ given]
∴By Side-Angel-Side criterion of congruence, we have
ΔAOD ≅ ΔCOD
(iii) The corresponding parts of the congruent
triangles are congruent.
∴AD = CD ...[c. p.c t ]
Hence proved.
The following figure has 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.
Prove that: (i) AM = AN (ii) ΔAMC ≅ ΔANB
In ΔABC, AB = AC. m and N are points on
AB and AC such that BM = CN
BN and CM are joined
(i) In ΔAMC and ΔANB
AB = AC ...[ Given ] ...(1)
BM = CN ....[ Given ] ...(2)
Subtracting (2) from (1), we have
AB - BM = AC - CN
⇒ AM = AN ...(3)
(ii) Consider the triangles AMC and ANB
AC = AB ...[ given ]
∠AMC = ∠ANB ...[ common 90° ]
AM = AN ....[ from ( 3 ) ]
∴ By the Side-Angel-Side Criterion of congruence, we have ΔAMC ≅ ΔANB
The following figure has 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.
Prove that: (i) BN = CM (ii) ΔBMC≅ΔCNB
In ΔABC, AB = AC. m and N are points on
AB and AC such that BM = CN
BN and CM are joined
(i) The corresponding parts of the congruent triangles are congruent.
∴ CM = BN ....[ c.p.c.t ] ...(1)
(ii) Consider the triangles ΔBMC and ΔCNB
BM = CN ...[ given ]
BC = BC ...[ common ]
Cm = BN ..[ from (1) ]
∴ By Side-Side-Side criterion of congruence, we have ΔBMC ≅ ΔCNB
In a triangle, ABC, AB = BC, AD is perpendicular to side BC and CE is perpendicular to side AB.
Prove that: AD = CE.
ln ΔABD and ΔCBE,
AB = BC ....( given )
∠ ADB = ∠ CEB = 90° ....[Perpendiculars]
∠B = ∠B ....( Common angle )
∴ ΔABD ≅ ΔCBE ....( by AAS congruence )
⇒ AD = CE ...( c.p.c.t. )
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.
Given: PL = RM
To prove: SP = PQ and MP = PL
Proof:
Since SR and PQ are opposite sides of a parallelogram,
PQ = SR ...(i)
Also, PL = RM ...(ii)
Subtracting (ii) from (i),
PQ - PL = SR - RM
⇒ LQ = SM ....(3)
Now, in ΔSMP and ΔQLP,
∠MSP = ∠PQL ....( alternate interior angles )
∠SMP = ∠PLQ ....( alternate interior angles )
SM = LQ ....[ from(3) ]
∴ ΔSMP ≅ ΔQLP ....( by ASA congruence )
⇒ SP = PQ and MP = PL ....( c.p.c.t. )
⇒ 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 up to 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 ).
ΔABC is an equilateral triangle,
So, each of its angles equals 60°.
QP is parallel to AC,
⇒ ∠PQB = ∠RAQ = 60°
ln ΔQBP,
∠PQB = ∠BQP = 60°
So, ∠PBQ + ∠BQP + ∠BPQ = 180° ....(angle sum property)
⇒ 60°+ 60° + ∠BPQ = 180°
⇒ ∠BPQ = 60°
So, ΔBPQ is an equilateral triangle.
⇒ QP = BP
⇒ QP = CR ....(i)
Now, ∠QPM + ∠BPQ = 180° ...(linear pair)
⇒ ∠QPM+ 60°= 180°
⇒ ∠QPM = 120°
Also, ∠RCM+ ∠ACB = 180° ...(linear pair)
⇒ ∠RCM+ 60° = 180°
⇒ ∠RCM = 120°
ln ΔRCM and ΔQMP,
∠RCM = ∠QPM ....(each is 120°)
∠RMC = ∠QMP ...(vertically opposite angles)
QP= CR ....(from(i))
⇒ ΔRCM ≅ ΔQMP ....(AAS congruence criterion)
So, CM = PM
⇒ QR bisects PC.
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.
Sol:
In ΔAOD and ΔBOC,
∠ AOD = ∠ BOC ....(vertically opposite angles)
∠ DAO = ∠ CBO ....(each 90°)
AD = BC ....(given)
∴ ΔAOD ≅ ΔBOC ...(by AAS congruence criterion)
⇒ AO = BO ...(c.p.c.t.)
⇒ 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.
In ΔABC,
AB = AC
⇒ ∠B = ∠C ...( angles opposite to equal sides are equal )
⇒
⇒ ∠OBC = ∠OCB ...[ ∵ OB and OC are bisectors of ∠B and ∠C respectively, ∠OBC =
⇒ OB = OC ...( Sides opposite to equal angles are equal ) ...(ii)
Now, in ΔABO and ΔACO,
AB = AC ...( given )
∠OBC = ∠OCB ...[ from(i) ]
OB = OC ...[ from(ii) ] ...( proved )
∴ ΔABO ≅ ΔACO ...( by SAS congruence criterion )
⇒ ∠BAO = ∠CAO ...( c.p.c.t. )
⇒ AO bisects ∠BAC ...(proved)
In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.
Prove that AD = FC.
Given that, BC = DE
⇒ BC + CD = DE + CD ....( Adding CD on both sides )
⇒ BD = CE ....(i)
Now, in ΔABD and ΔFEC,
AB = EF ....(given)
∠ABD = ∠FEC ....(Each 90°)
BD = CE ....[ From (i) ]
⇒ ΔABD ≅ ΔFEC ...(by SAS congruence criterion)
⇒ AD = FC ...(c.p.c.t.)
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.
Sol:
In ΔAOD and ΔAOB,
AD = AB ...(given)
AO = AO ...(Common)
OD = OB ...(given)
⇒ ΔAOD ≅ ΔAOB ...(by SSS congruence criterion)
⇒ ∠AOD = ∠AOB ...(c.p.c.t.) ...(i)
Similarly, ΔDOC ≅ ΔBOC
⇒ ∠DOC = ∠BOC ...(c.p.c.t.) ...(ii)
But, ∠AOB + ∠AOD + ∠COD + ∠BOC = 4 Right angles ...[ Sum of the angles at a point is 4 Right angles ]
⇒ 2∠AOD + 2∠COD = 4 Right angles ....[ Using (i) and (ii) ]
⇒ ∠AOD + ∠COD = 2 Right angles
⇒ ∠AOD + ∠COD = 180°
⇒ ∠AOD and ∠COD form a linear pair.
⇒ AO and OC are in the same straight line.
⇒ AOC is a straight line.
In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
Given: In quadrilateral ABCD, AD = BC and BD = AC.
To Prove:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
Proof:
In ΔABD and ΔBAC,
AD = BC ....(given)
BD = CA ....(given)
AB = AB ....(common)
∴ ΔABD ≅ ΔBAC ....(by SSS congruence criterion)
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