From the following figure, prove that: AB > CD.
In ΔABC,
AB = AC ...[ Given ]
∴ ∠ACB = ∠B ...[ angles opposite to equal sides are equal ]
∠B = 70° ...[ Given ]
⇒ ∠ACB = 70° ...(i)
Now,
∠ACB +∠ACD = 180°...[ BCD is a straight line]
⇒ 70° + ∠ACD = 180°
⇒ ∠ACD = 110° ...(ii)
In ΔACD,
∠CAD + ∠ACD + ∠D = 180°
⇒ ∠CAD + 110° + ∠D = 180° ...[ From (ii) ]
⇒ ∠CAD + ∠D = 70°
But ∠D = 40° ...[ Given ]
⇒ ∠CAD + 40°= 70°
⇒ ∠CAD = 30° …(iii)
In ΔACD,
∠ACD = 110° ...[ From (ii) ]
∠CAD = 30° ...[ From (iii) ]
∠D = 40° ...[ Given ]
∴ D > ∠CAD
⇒ AC > CD ....[Greater angle has greater side opposite to it]
Also,
AB = AC ...[ Given ]
Therefore, AB > CD.
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
Sol:
In ΔPQR,
QR = PR ...[ Given ]
∴ ∠P = ∠Q ...[ angles opposite to equal sides are equal ]
⇒ ∠P = 36° ..[Given]
⇒ ∠Q = 36°
In ΔPQR,
∠P + ∠Q + ∠R = 180°
⇒ 36° + 36° + ∠R = 180°
⇒ ∠R + 72° = 180°
⇒ ∠R = 108°
Now,
∠R = 108°
∠P = 36°
∠Q = 36°
Since ∠R is the greatest, therefore, PQ is the largest side.
If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.
The sum of any two sides of the triangle is always greater than third side of the triangle.
Third side < 13 + 8 = 21 cm.
The difference between any two sides of the triangle is always less than the third side of the triangle.
Third side > 13 - 8 = 5 cm.
Therefore, the length of the third side is between 5 cm and 9 cm, respectively.
The value of a = 5 cm and b = 21cm.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
In ΔABC,
AB = AC
⇒ ∠ABC = ∠ACB ..( angles opposite to equal sides are equal )
⇒ ∠ABC = ∠ACB = 67°
⇒ ∠BAC = 180° - ∠ABC - ∠ACB ..( angle sum property )
⇒ ∠BAC = 180° - 67° - 67° = 46°
Since ∠BAC < ∠ABC, we have
BC< AC ...( 1 )
Now, ∠ACD = 180° - ACB ...( liner pair )
⇒ ∠ACD = 180° - 67° = 113°
Thus, in ΔACD,
∠CAD = 180°- ∠ACD - ∠ADC
⇒ ∠CAD = 180° - 113° - 33° = 34°
Since ∠ADC < ∠CAD, we have
AC < CD ...( 2 )
From ( 1 ) and ( 2 ), we have
BC < AC < CD.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
In ΔABC,
∠BAC < ∠ABC
BC < AC ....( 1 )
Now, ∠ACB = 180° - ∠ABC - ∠BAC
∠ACB = 180° - 73° - 47°
∠ACB = 60°
Now, ∠ACD = 180° - ∠ACB
∠ACD = 180° - 60° = 120°
Now, in ΔACD,
∠ADC = 180° - ∠ACD - ∠CAD
∠ADC = 180° - 120° - 31°
∠ADC = 29°
Since ∠ADC < ∠CAD, we have
AC < CD ....( 2 )
From ( 1 ) and ( 2 ), we have
BC < AC < CD.
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
∠BAC = 180° - ∠BAD = 180° - 137° = 43°
∠ABC = 180° - ∠ABE = 180° - 106° = 74°
Thus, in ΔABC,
∠ACB = 180° - ∠BAC - ∠ABC
⇒ ∠ACB = 180° - 43° - 74° = 63°
Now, ∠ABC = ∠OBC + ∠ABO
⇒ ∠ABC = 2∠OBC ....( OB is biosector of ∠ABC )
⇒ 74° = 2∠OBC
⇒ ∠OBC = 37°
Similarly,
∠ACB = ∠OCB + ∠ACO
⇒ ∠ACB = 2∠OCB ...( OC is bisector of ACB )
⇒ 63° = 2∠OCB
⇒ ∠OCB = 31.5°
Now, in ΔBOC,
∠BOC = 180° - ∠OBC - ∠OCB
⇒ ∠BOC = 180° - 37° - 31.5°
⇒ ∠BOC = 111.5°
Since, ∠BOC> ∠OBC > ∠OCB, we have
BC > OC > OB
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
AD > AC ...( Given )
⇒ ∠C > ∠ADC ...( 1 )
Now, ∠ADC > ∠B + ∠BAC ...( Exterior Angel Property )
⇒ ∠ADC > ∠B ....( 2 )
From ( 1 ) and ( 2 ), we have
⇒ ∠C > ∠ADC > ∠B
⇒ ∠C > ∠B
⇒ AB > AC
In the following figure, ∠BAC = 60o and ∠ABC = 65o.
Prove that:
(i) CF > AF
(ii) DC > DF
In ΔBEC,
∠B + ∠BEC + ∠BCE = 180°
∠B = 65° ...[Given]
∠BEC = 90° ...[CE is perpendicular to AB]
⇒ 65° + 90° + ∠BCE = 180°
⇒ ∠BCE = 180° - 155°
⇒ ∠BCE = 25°= ∠DCF …(i)
In ΔCDF,
∠DCF + ∠FDC + ∠CFD = 180°
∠DCF = 25° ....[From (i)]
∠FDC = 90° ...[ AD is perpendicular to BC]
⇒ 25°+ 90°+ ∠CFD = 180°
⇒ ∠CFD = 180° - 115°
⇒ ∠CFD = 65° …(ii)
Now, ∠AFC + ∠CFD = 180° ....[AFD is a straight line]
⇒ ∠AFC + 65° = 180°
⇒ ∠AFC = 115° …(iii)
In ΔACE,
∠ACE + ∠CEA + ∠BAC = 180°
∠BAC = 60° ....[Given]
⇒ ∠CEA = 90° ...[CE is perpendicular to AB]
⇒ ∠ACE + 90° + 60° = 180°
⇒ ∠ACE = 180° - 150°
∠ACE = 30° …(iv)
In ΔAFC,
∠AFC + ∠ACF + ∠FAC = 180°
∠AFC = 115° ....[From (iii)]
∠ACF = 30° ...[From (iv)]
⇒ 115° + 30° + ∠FAC = 180°
⇒ ∠FAC = 180° - 145°
⇒ ∠FAC = 35° …(v)
In ΔAFC,
⇒ ∠FAC = 35° ...[ From (v) ]
⇒ ∠ACF = 30° ...[ From (iv) ]
∴ ∠FAC > ∠ACF
⇒ CF > AF
In Δ CDF,
∠DCF = 25° ...[From (i)]
∠CFD = 65° ...[From (ii)]
∴ ∠CFD > ∠DCF
⇒ DC > DF
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
∠ACB = 74° ...(i)[ Given ]
∠ACB + ∠ACD = 180° ....[ BCD is a straight line ]
⇒ 74° + ∠ACD = 180°
⇒ ∠ACD = 106° …..(ii)
In ΔACD,
∠ACD + ∠ADC+ ∠CAD = 180°
Given that AC = CD
⇒ ∠ADC= ∠CAD
⇒ 106° + ∠CAD + ∠CAD = 180° ....[From (ii)]
⇒ 2∠CAD = 74°
⇒ ∠CAD = 37° = ∠ADC ...(iii)
Now,
∠BAD = 110° ....[Given]
∠BAC + ∠CAD = 110°
∠BAC + 37° = 110°
∠ BAC = 73° ….(iv)
In ABC,
∠ B + ∠BAC + ∠ACB = 180°
∠B + 73° + 74° = 180° ...[From (i) and (iv)]
∠B + 147°= 180°
∠B = 33° …..(v)
∴ ∠BAC > ∠B ...[ From (iv) and (v)]
⇒ BC > AC
But,
AC = CD ...[ Given ]
⇒ BC > CD
From the following figure; 
prove that:
(i) AB > BD
(ii) AC > CD
(iii) AB + AC > BC.
(i) ∠ADC + ∠ADB = 180° ...[ BDC is a straight line ]
∠ADC = 90° ...[ Given ]
90° + ∠ADB = 180°
∠ADB = 90° ....(i)
In ΔADB,
∠ADB = 90° ....[ From (i) ]
∴ ∠B + ∠BAD = 90°
Therefore, ∠B and ∠BAD are both acute, that is less than 90°.
∴ AB > BD ….(ii)[ Side opposite 90° angle is greater than the side opposite acute angle ]
(ii) In ΔADC,
∠ADB = 90°
∴ ∠C + ∠DAC = 90°
Therefore, ∠C and ∠DAC are both acute, which is less than 90°.
∴ AC > CD ...(iii)[ Side opposite 90° angle is greater than side opposite acute angle ]
Adding (ii) and (iii)
AB + AC > BD + CD
⇒ AB + AC > BC
In a quadrilateral ABCD; prove that:
(i) AB+ BC + CD > DA
(ii) AB + BC + CD + DA > 2AC
(iii) AB + BC + CD + DA > 2BD
Const: Join AC and BD.
(i) In ΔABC,
AB + BC > AC ….(i)[ Sum of two sides is greater than the third side ]
In ΔACD,
AC + CD > DA ….(ii)[ Sum of two sides is greater than the third side ]
Adding (i) and (ii)
AB + BC + AC + CD > AC + DA
AB + BC + CD > AC + DA - AC
AB + BC + CD > DA ….(iii)
(ii) In ΔACD,
CD + DA > AC ….(iv) [Sum of two sides is greater than the third side]
Adding (i) and (iv)
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
(iii) In ΔABD,
AB + DA > BD ….(v)[Sum of two sides is greater than the third side]
In ΔBCD,
BC + CD > BD ….(vi)[Sum of two sides is greater than the third side]
Adding (v) and (vi)
AB + DA + BC + CD > BD + BD
AB + DA + BC + CD > 2BD
In the following figure, ABC is an equilateral triangle and P is any point in AC;
prove that: BP > PA
In ΔABC,
AB = BC = CA ...[ ABC is an equilateral triangle ]
∴ ∠A = ∠B = ∠C
∴ ∠A = ∠B = ∠C =
In ΔABP,
∠A = 60°
∠ABP< 60°
∴ ∠A > ∠ABP
⇒ BP > PA ....[ Side opposite to greater side is greater ]
In the following figure, ABC is an equilateral triangle and P is any point in AC;
prove that: BP > PC
In ΔBPC,
∠C = 60°
∠CBP < 60°
∴∠C > ∠CBP
⇒ BP > PC ....[ Side opposite to greater side is greater ]
P is any point inside the triangle ABC.
Prove that: ∠BPC > ∠BAC.
Let ∠PBC = x and ∠PCB = y
then,
∠BPC = 180° - ( x + y ) …(i)
Let ∠ABP = a and ∠ACP = b
then,
∠BAC = 180° - ( x + a ) - ( y + b )
⇒ ∠BAC = 180° - ( x + y ) - ( a + b )
⇒ ∠BAC = ∠BPC - ( a + b )
⇒ ∠BPC = ∠BAC + ( a + b )
⇒ ∠BPC > ∠BAC.
Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle.
Sol:
We know that the exterior angle of a triangle is always greater than each of the interior opposite angles.
∴ In ΔABD,
∠ADC > ∠B ...(i)
In ΔABC,
AB = AC
∴∠B = ∠C ...(ii)
From (i) and (ii)
∠ADC > ∠C
(i) In ΔADC,
∠ADC > ∠C
∴AC > AD ....(iii)[ side opposite to greater angle is greater ]
(ii) In ΔABC,
AB = AC
⇒ AB > AD ...[ From (iii) ]
In the following diagram; AD = AB and AE bisect angle A. 
Prove that:
(i) BE = DE
(ii) ∠ABD > ∠C
Const: Join ED.
In ΔAOB and ΔAOD,
AB = AD ...[ Given ]
AO = AO ....[ Common ]
∠BAO = ∠DAO ....[ AO is bisector of A ]
∴ ΔAOB ≅ ΔAOD ....[ SAS criterion ]
Hence,
BO = OD …(i)[ c.p.c.t. ]
∠AOB = ∠AOD …(ii)[ c.p.c.t. ]
∠ABO = ∠ADO ⇒ ∠ABD = ∠ADB …(iii)[ c.p.c.t. ]
Now,
∠AOB = ∠DOE ...[Vertically opposite angles]
∠AOD = ∠BOE ...[Vertically opposite angles]
∠BOE = ∠DOE …(iv)[ From (ii) ]
(i) In ΔBOE and ΔDOE,
BO = CD ...[ From (i) ]
OE = OE ...[ Common ]
∠BOE = ∠DOE ...[ From (iv) ][ SAS criterion ]
Hence, BE = DE ...[ c.p.c.t. ]
(ii) In BCD,
∠ADB = ∠C + ∠CBD ...[ Ext. angle = sum of opp. int. angles ]
⇒ ∠ADB > ∠C
⇒ ∠ABD > ∠C ...[ From (iii) ]
Const: Join ED.
In ΔAOB and ΔAOD,
AB = AD ...[ Given ]
AO = AO ....[ Common ]
∠BAO = ∠DAO ....[ AO is bisector of A ]
∴ ΔAOB ≅ ΔAOD ....[ SAS criterion ]
Hence,
BO = OD …(i)[ c.p.c.t. ]
∠AOB = ∠AOD …(ii)[ c.p.c.t. ]
∠ABO = ∠ADO ⇒ ∠ABD = ∠ADB …(iii)[ c.p.c.t. ]
Now,
∠AOB = ∠DOE ...[Vertically opposite angles]
∠AOD = ∠BOE ...[Vertically opposite angles]
∠BOE = ∠DOE …(iv)[ From (ii) ]
(i) In ΔBOE and ΔDOE,
BO = CD ...[ From (i) ]
OE = OE ...[ Common ]
∠BOE = ∠DOE ...[ From (iv) ][ SAS criterion ]
Hence, BE = DE ...[ c.p.c.t. ]
(ii) In BCD,
∠ADB = ∠C + ∠CBD ...[ Ext. angle = sum of opp. int. angles ]
⇒ ∠ADB > ∠C
⇒ ∠ABD > ∠C ...[ From (iii) ]
In ΔABC,
AB > AC,
⇒ ∠ABC < ∠ACB
∴ 180° - ∠ABC > 180° - ∠ACB
⇒
⇒ 90° -
⇒ ∠CBP > ∠BCP ...[ BP is bisector of ∠CBD and CP is bisector of ∠BCE ]
⇒ PC > PB ...[ Side opposite to greater angle is greater ]
In the following figure; AB is the largest side and BC is the smallest side of triangle ABC.
Write the angles xo, yo and zo in ascending order of their values.
Since AB is the largest side and BC is the smallest side of the triangle ABC.
AB > AC > BC
⇒ 180° - z° > 180° - y° > 180° - x°
⇒ - z° > -y° > - x°
⇒ z° > y° > x°.
In quadrilateral ABCD, side AB is the longest and side DC is the shortest.
Prove that: C > A.
In the quadrilateral ABCD,
Since AB is the longest side and DC is the shortest side.
∠1 > ∠2 [ AB > BC ]
∠ 7 > ∠4 [ AD > DC ]
∴ ∠1 + ∠7 > ∠2 + ∠4
⇒ ∠C > ∠A.
In quadrilateral ABCD, side AB is the longest and side DC is the shortest.
Prove that: D > B.
∠5 > ∠6 [ AB > AD ]
∠3 > ∠8 [ BC > CD ]
∴ ∠5 + ∠3 > ∠6 + ∠8
⇒ ∠D > ∠B
In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D,
prove that: ∠ADC is greater than ∠ADB.
In ΔADC,
∠ADB = ∠1 + ∠C ....(i)
In ADB,
∠ADC = ∠2 + ∠B ...(ii)
But AC > AB ....[ Given ]
⇒ ∠B > ∠C
Also given, ∠2 = ∠1 ....[ AD is bisector of A ]
⇒ ∠2 + ∠B > ∠1 + ∠C ...(iii)
From (i), (ii) and (iii)
⇒ ∠ADC > ∠ADB.
In an isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that:
(i) AC > AD
(ii) AE > AC
(iii) AE > AD
We know that the bisector of the angle at the vertex of an isosceles triangle bisects the base at the right angle.
Using Pythagoras theorem in AFB,
AB2 = AF2 + BF2 ...(i)
In AFD,
AD2 = AF2 + DF2 ...(ii)
We know ABC is isosceles triangle and AB = AC
AC2 = AF2 + BF2 ..(iii)[ From (i)]
Subtracting (ii) from (iii)
AC2 - AD2 = AF2 + BF2 - AF2 - DF2
AC2 - AD2 = BF2 - DF2
Let 2DF = BF
AC2 - AD2 = (2DF)2 - DF2
AC2 - AD2 = 4DF2 - DF2
AC2 = AD2 + 3DF2
⇒ AC2 > AD2
⇒ AC > AD
Similarly, AE > AC and AE > AD.
Given: ED = EC
Prove: AB + AD > BC.
The sum of any two sides of the triangle is always greater than the third side of the triangle.
ln ΔCEB,
CF + EB > BC
⇒ DE + EB > BC ...[ CE = DE ]
⇒ DB > BC ...(i)
ln ΔADB,
AD+ AB > BD
⇒ AD + AB > BD > BC ...[ from(i) ]
⇒ AD+ AB > BC
In triangle ABC, AB > AC and D is a point inside BC.
Show that: AB > AD.
Given that, AB > AC
⇒ ∠C > ∠B ...(i)
Also in ΔADC
∠ADB = ∠DAC + ∠C ...[ Exterior angle ]
⇒ ∠ADB > ∠C
⇒ ∠ADB > ∠C > ∠B ...[ From(i) ]
⇒ ∠ADB > ∠B
⇒ AB > AD.
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