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RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.
In ∆ABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
In ∆ABC, ∠A = 40°, ∠B = 60°
But ∠A + ∠B + ∠C = 180°

⇒ 40° + 60° + ∠C = 180°
⇒ ∠C = 180° = (40° + 60°)
= 180° – 100° = 80°
∵ ∠C = 80°, which is the greatest angle and
∠A = 40° is the smallest angle
∴ Side AB which is opposite to the greatest angle is the longest and side BC which is opposite to the smallest angle is the shortest.

Question 2.
In a ∆ABC, if ∠B = ∠C = 45°. which is the longest side?
Solution:
In ∆ABC, ∠B = ∠C = 45°
But ∠A + ∠B + ∠C = 180°

⇒ ∠A + 45° + 45° = 180°
⇒ ∠A + 90° = 180°
∴ ∠A = 180°-90° = 90°
∴∠A is the greatest
∴ Side BC opposite to it is the longest

Question 3.
In ∆ABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°, prove that :
(i) AD > CD
(ii) AD > AC
Solution:
Given : In AABC, side BC is produced to D such that BD = BC
∠A = 70° and ∠B = 60°

To prove :
(i) AD > CD (ii) AD > AC
Proof: In ∆ABC,
∠A = 70°, ∠B = 60°
But Ext. ∠CBD + ∠CBA = 180° (Linear pair)
∠CBD + 60° = 180° 3
⇒ ∠CBD = 180° – 60° = 120°
But in ∆BCD,
BD = BC
∴ ∠D = ∠BCD
But ∠D + ∠BCD = 180° – 120° = 60°
∴∠D = ∠BCD = \(\frac { { 60 }^{ \circ } }{ 2 }\)  = 30°
and in ∆ABC,
∠A + ∠B + ∠C = 180°
⇒ 70° + 60° + ∠C = 180°
⇒ 130° + ∠C = 180°
∴ ∠C =180°- 130° = 50°
Now ∠ACD = ∠ACB + ∠BCD = 50° + 30° = 80°
(i) Now in ∆ACB,
∠ACD = 80° and ∠A = 70°
∴ Side AD > CD
(Greater angle has greatest side opposite to it)
(ii) ∵ ∠ACD = 80° and ∠D = 30°
∴ AD > AC

Question 4.
Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Solution:
We know that in a triangle, sum of any two sides is greater than the third side and 2 cm + 3 cm = 5 cm and 5 cm < 7 cm
∴ This triangle is not possible to draw

Question 5.
In ∆ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
Solution:
In ∆ABC, ∠B = 35°, ∠C = 65° and AP is the bisector of ∠BAC which meets BC in P.
Arrange PA, PB and PC in descending order In ∆ABC,
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ ∠A + 35° + 65° = 180°
∠A + 100°= 180°
∴ ∠A =180°- 100° = 80°
∵ PA is a bisector of ∠BAC
∴ ∠1 = ∠2 = \(\frac { { 80 }^{ \circ } }{ 2 }\)  = 40°
Now in ∆ACP, ∠ACP > ∠CAP
⇒ ∠C > ∠2
∴ AP > CP …(i)
Similarly, in ∆ABP,
∠BAP > ∠ABP ⇒ ∠1 > ∠B
∴ BP > AP …(ii)
From (i) and (ii)
BP > AP > CP

Question 6.
Prove that the perimeter of a triangle is greater than the sum of its altitudes
Solution:
Given : In ∆ABC,
AD, BE and CF are altitudes

To prove : AB + BC + CA > AD + BC + CF
Proof : We know that side opposite to greater angle is greater.
In ∆ABD, ∠D = 90°
∴ ∠D > ∠B
∴ AB >AD …(i)
Similarly, we can prove that
BC > BE and
CA > CF
Adding we get,
AB + BC + CA > AD + BE + CF

Question 7.
In the figure, prove that:
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC

Solution:
Given : In the figure, ABCD is a quadrilateral and AC is joined
To prove :
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
Proof:
(i) In ∆ABC,
AB + BC > AC …(i)
(Sum of two sides of a triangle is greater than its third side)
Similarly in ∆ADC,
CD + DA > AC …(ii)
Adding (i) and (ii)
CD + DA + AB + BC > AC + AC
⇒ CD + DA + AB + BC > 2AC
(ii) In ∆ACD,
CD + DA > CA
(Sum of two sides of a triangle is greater than its third side)
Adding AB to both sides,
CD + DA + AB > CA + AB
But CA + AB > BC (in ∆ABC)
∴ CD + DA + AD > BC

Question 8.
Which of the following statements are true (T) and which are false (F)?
(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.
(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.
(iii) Sum of any two sides of a triangle is greater than the third side.
(iv) Difference of any two sides of a triangle is equal to the third side.
(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.
(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.
Solution:
(i) False. Sum of three sides of a triangle is greater than the sum of its altitudes.
(ii) True.
(iii) True.
(iv) False. Difference of any two sides is less than the third side.
(v) True.
(vi) True.

Question 9.
Fill in the blanks to make the following statements true.
(i) In a right triangle, the hypotenuse is the ……. side.
(ii) The sum of three altitudes of a triangle is ……. than its perimeter.
(iii) The sum of any two sides of a triangle is …….. than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the ….. side opposite to it.
(v) Difference of any two sides of a triangle is……. than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has ……… angle opposite to it.
Solution:
(i) In a right triangle, the hypotenuse is the longest side.
(ii) The sum of three altitudes of a triangle is less than its perimeter.
(iii) The sum of any two sides of a triangle is greater than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.
(v) Difference of any two sides of a triangle is less than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has greater angle opposite to it.

Question 10.
O is any point in the interior of ∆ABC. Prove that
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > \(\frac { 1 }{ 2 }\) (AB + BC + CA)
Solution:
Given : In ∆ABC, O is any point in the interior of the ∆ABC, OA, OB and OC are joined
To prove :
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > \(\frac { 1 }{ 2 }\) (AB + BC + CA)
Construction : Produce BO to meet AC in D.

Proof: In ∆ABD,
(i) AB + AD > BD (Sum of any two sides of a triangle is greater than third)
⇒ AB + AD > BO + OD …(i)
Similarly, in ∆ODC,
OD + DC > OC …(ii)
Adding (i) and (ii)
AB + AD + OD + DC > OB + OD + OC
⇒ AB + AD + DC > OB + OC
⇒ AB + AC > OB + OC
(ii) Similarly, we can prove that
BC + AB > OA + OC
and CA + BC > OA + OB
(iii) In ∆OAB, AOBC and ∆OCA,
OA + OB > AB
OB + OC > BC
and OC + OA > CA
Adding, we get
2(OA + OB + OC) > AB + BC + CA
∴ OA + OB + OO > \(\frac { 1 }{ 2 }\) (AB + BC + CA)

Question 11.
Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.
Solution:
Given : In quadrilateral ABCD, AC and BD are its diagonals,

To prove : AB + BC + CD + DA > AC + BD
Proof: In ∆ABC,
AB + BC > AC …(i)
(Sum of any two sides of a triangle is greater than its third side)
Similarly, in ∆ADC,
DA + CD > AC …(ii)
In ∆ABD,
AB + DA > BD …(iii)
In ∆BCD,
BC + CD > BD …(iv)
Adding (i), (ii), (iii) and (iv)
2(AB + BC + CD + DA) > 2AC + 2BD
⇒ 2(AB + BC + CD + DA) > 2(AC + BD)
∴ AB + BC + CD + DA > AC + BD

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.
ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.
Solution:
Given : In ∆ABC, D is mid-point of BC and DE ⊥ AB, DF ⊥ AC and DE = DF

To Prove : ∆ABC is an isosceles triangle
Proof : In right ∆BDE and ∆CDF,
Side DE = DF
Hyp. BD = CD
∴ ∆BDE ≅ ∆CDF (RHS axiom)
∴ ∠B = ∠C (c.p.c.t.)
Now in ∆ABC,
∠B = ∠C (Prove)
∴ AC = AB (Sides opposite to equal angles)
∴ AABC is an isosceles triangle

Question 2.
ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that ∆ABC is an isosceles.
Solution:
Given : In ∆ABC,
BE ⊥ AC and CF ⊥ AB
BE = CF

To prove : AABC is an isosceles triangle
Proof : In right ABCE and ABCF Side
BE = CF (Given)
Hyp. BC = BC (Common)
∴ ∆BCE ≅ ∆BCF (RHS axiom)
∴ ∠BCE = ∠CBF (c.p.c.t.)
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle

Question 3.
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
Solution:
Given : A point P lies in the angle ABC and PL ⊥ BA and PM ⊥ BC and PL = PM. PB is joined

To prove : PB is the bisector ∠ABC,
Proof : In right ∆PLB and ∆PMB
Side PL = PM (Given)
Hyp. PB = PB (Common)
∴ ∆PLB ≅ ∆PMB (RHS axiom)
∴ ∆PBL = ∆PBM (c.p.c.t.)
∴ PB is the bisector of ∠ABC

Question 4.
In the figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.

Solution:
Given : In the figure,
AD ⊥ CD and CB ⊥ CD, AQ = BP and DP = CQ

To prove : ∠DAQ = ∠CBP
Proof : ∵ DP = CQ
∴ DP + PQ = PQ + QC
⇒ DQ = PC
Now in right ∆ADQ and ∆BCP
Side DQ = PC (Proved)
Hyp. AQ = BP
∴ ∆ADQ ≅ ∆BCP (RHS axiom)
∴ ∠DAQ = ∠CBP (c.p.c.t.)

Question 5.
Which of the following statements are true (T) and which are false (F):
(i) Sides opposite to equal angles of a triangle may be unequal.
(ii) Angles opposite to equal sides of a triangle are equal.
(iii) The measure of each angle of an equilateral triangle is 60°.
(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.
(v) The bisectors of two equal angles of a triangle are equal.
(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.
(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.
(viii)If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.
Solution:
(i) False : Sides opposite to equal angles of a triangle are equal.
(ii) True.
(iii) True.
(iv) False : The triangle is an isosceles triangle.
(v) True.
(vi) False : The triangle is an isosceles.
(vii) False : The altitude an equal.
(viii) False : If one side and hypotenuse of one right triangle on one side and hypotenuse of the other right triangle are equal, then triangles are congruent.
(ix) True.

Question 6.
Fill in the blanks in the following so that each of the following statements is true.
(i) Sides opposite to equal angles of a triangle are …….
(ii) Angle opposite to equal sides of a triangle are …….
(iii) In an equilateral triangle all angles are …….
(iv) In a ∆ABC if ∠A = ∠C, then AB = …….
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = ……..
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ……… CE.
(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ∆ABC ≅ ∆……
Solution:
(i) Sides opposite to equal angles of a triangle are equal.
(ii) Angle opposite to equal sides of a triangle are equal.
(iii) In an equilateral triangle all angles are equal.
(iv) In a ∆ABC, if ∠A = ∠C, then AB = BC.
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = AC.
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is equal to CE.
(vii) In right triangles ABC and DEF, it hypotenuse AB = EF and side AC = DE, then ∆ABC ≅ ∆EFD.

Question 7.
ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
Solution:
Given : In square ABCD, X and Y are points on side AD and BC respectively and AY = BX

To prove : BY = AX
∠BAY = ∠ABX
Proof: In right ∆BAX and ∆ABY
AB =AB (Common)
Hyp. BX = AY (Given)
∴ ∆BAX ≅ ∆ABY (RHS axiom)
∴ AX = BY (c.p.c.t.)
∠ABX = ∠BAY (c.p.c.t.)
Hence, BY = AX and ∠BAY = ∠ABX.

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.
In the figure, it is given that AB = CD and AD = BC. Prove that ∆ADC ≅ ∆CBA.

Solution:
Given : In the figure, AB = CD, AD = BC

To prove : ∆ADC = ∆CBA
Proof : In ∆ADC and ∆CBA
CD = AB (Given)
AD = BC (Given)
CA = CA (Common)
∴ ∆ADC ≅ ∆CBA (SSS axiom)

Question 2.
In a APQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Solution:
Given : In ∆PQR, PQ = QR
L, M and N are the mid-points of sides PQ, QR and RP respectively. Join LM, MN and LN

To prove : ∠PNM = ∠PLM
Proof : In ∆PQR,
∵ M and N are the mid points of sides PR and QR respectively
∴ MN || PQ and MN = \(\frac { 1 }{ 2 }\) PQ …(i)
∴ MN = PL
Similarly, we can prove that
LM = PN
Now in ∆NML and ∆LPN
MN = PL (Proved)
LM = PN (Proved)
LN = LN (Common)
∴ ∆NML = ∆LPN (SSS axiom)
∴ ∠MNL = ∠PLN (c.p.c.t.)
and ∠MLN = ∠LNP (c.p.c.t.)
⇒ ∠MNL = ∠LNP = ∠PLM = ∠MLN
⇒ ∠PNM = ∠PLM

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3

Other Exercises

• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
• RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Solution:
Given : In ∆ABC and ∆DEF,
∠B = ∠E = 90°
∠C = ∠F
AB = DE

To prove : ∆ABC = ∆DEF
Proof : In ∆ABC and ∆DEF,
∠B = ∠E (Each = 90°)
∠C = ∠F (Given)
AB = DE (Given)
∆ABC = ∆DEF (AAS axiom)

Question 2.
If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Solution:
Given : In ∆ABC, AE is the bisector of vertical exterior ∠A and AE \(\parallel\) BC

To prove : ∆ABC is an isosceles
Proof: ∵ AE \(\parallel\) BC
∴ ∠1 = ∠B (Corresponding angles)
∠2 = ∠C (Alternate angle)
But ∠1 = ∠2 (∵ AE is the bisector of ∠CAD)
∴ ∠B = ∠C
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle

Question 3.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Solution:
Given : In ∆ABC, AB = AC
∠A = 2(∠B + ∠C)

To calculate: Base angles,
Let ∠B = ∠C = x
Then ∠A = 2(∠B + ∠C)
= 2(x + x) = 2 x 2x = 4x
∵ Sum of angles of a triangle = 180°
∴ 4x + x + x – 180° ⇒ 6x = 180°
⇒ x= \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30° o
∴ ∠B = ∠C = 30 and ∠A = 4 x 30° = 120

Question 4.
Prove that each angle of an equilateral triangle is 60°. [NCERT]
Solution:
Given : ∆ABC is an equilateral triangle

Proof: In ∆ABC,
AB = AC (Sides of an equilateral triangle)
∴ ∠C = ∠B …(i)
(Angles opposite to equal angles)
Similarly, AB = BC
∴ ∠C = ∠A …(ii)
From (i) and (ii),
∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠A = ∠B = ∠C = \(\frac { { 180 }^{ \circ } }{ 3 }\)= 60°

Question 5.
Angles A, B, C of a triangle ABC are equal to each other. Prove that ∆ABC is equilateral.
Solution:
Given : In ∆ABC, ∠A = ∠B = ∠C

To prove : ∆ABC is an equilateral
Proof: In ∆ABC,
∴ ∠B = ∠C (Given)
∴ AC = AB …(i) (Sides opposite to equal angles)
Similarly, ∠C = ∠A
∴ BC =AB …(ii)
From (i) and (ii)
AB = BC = CA
Hence ∆ABC is an equilateral triangle

Question 6.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆ABC, ∠A = 90°

AB =AC (Given)
∴ ∠C = ∠B (Angles opposite to equal sides)
But ∠B + ∠C = 90° (∵ ∠B = 90°)
∴ ∠B = ∠C = \(\frac { { 90 }^{ \circ } }{ 2 }\) = 45°
Hence ∠B = ∠C = 45°

Question 7.
PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Solution:
Given : In ∆PQR, PQ = PR
S is a point on PQ and PT || QR

To prove : PS = PT
Proof : ∵ST || QR
∴ ∠S = ∠Q and ∠T = ∠R (Corresponding angles)
But ∠Q = ∠R (∵ PQ = PR)
∴ PS = PT (Sides opposite to equal angles)

Question 8.
In a ∆ABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.
Solution:
Given : In ∆ABC, AB = AC the bisectors of ∠B and ∠C intersect at O. M is any point on BO produced.

To prove : ∠MOC = ∠ABC
Proof: In ∆ABC, AB = BC
∴ ∠C = ∠B
∵ OB and OC are the bisectors of ∠B and ∠C
∴ ∠1 =∠2 = \(\frac { 1 }{ 2 }\)∠B
Now in ∠OBC,
Ext. ∠MOC = Interior opposite angles ∠1 + ∠2
= ∠1 + ∠1 = 2∠1 = ∠B
Hence ∠MOC = ∠ABC

Question 9.
P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
Solution:
Given : In ∆ABC, P is a point on the bisector of ∠B and from P, RPQ || AB is draw which meets BC in Q

To prove : ∆BPQ is an isosceles
Proof : ∵ BD is the bisectors of CB
∴ ∠1 = ∠2
∵ RPQ || AB
∴ ∠1 = ∠3 (Alternate angles)
But ∠1 == ∠2 (Proved)
∴ ∠2 = ∠3
∴ PQ = BQ (sides opposite to equal angles)
∴ ∆BPQ is an isosceles

Question 10.
ABC is a triangle in which ∠B = 2∠C, D is a point on BC such that AD bisects ∠BAC = 72°.
Solution:
Given: In ∆ABC,
∠B = 2∠C, AD is the bisector of ∠BAC AB = CD
To prove : ∠BAC = 72°
Construction : Draw bisector of ∠B which meets AD at O

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.