NCERT Solutions for Class 9 Mathematics Chapter 7 Triangles
Consequently, NCERT resourceful solutions of ORCHIDS to the students of Class 9 of Mathematics, Chapter 7: "Triangles," makes them able to develop real good improvement in understanding, which will make them perform well in examinations, and it enables students to become confident about the approach toward studying mathematics.
Access Answers to NCERT Solutions for Class 9 Mathematics Chapter 7 Triangles
Students can access the NCERT Solutions for Class 9 Mathematics Chapter 7 Triangles. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 7.1
In quadrilateral ABCD (See figure). AC = AD and AB bisects 
Given: In quadrilateral ABCD, AC = AD and AB bisects 
To prove: 
Proof: In 
AC = AD [Given]
AB = AB [Common]
Thus BC = BD [By C.P.C.T.]
ABCD is a quadrilateral in which AD = BC and 
(i) 
(ii) BD = AC
(iii) 
(i)In 
BC = AD [Given]
AB = AB [Common]
Thus AC = BD [By C.P.C.T.]
(ii)Since 
(iii)Since 
AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB (See figure)
In 
BC = AD [Given]
AC being a transversal. [Given]
Therefore 
Now 
And AC being a transversal. [Given]
Therefore 
Now In 
AC = AC [Common]
. Line 
(i) 
(ii) BP = BQ or P is equidistant from the arms of 
Given: Line 
(i) In 
AB = AB [Common]
(ii) Since 
In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC
To prove:
The line segment BC and DE are similar i.e. BC = DE
Proof:
We know that ∠BAD = ∠EAC
Now, by adding ∠DAC on both sides we get,
∠BAD + ∠DAC = ∠EAC +∠DAC
This implies, ∠BAC = ∠EAD
Now, ΔABC and ΔADE are similar by SAS congruency since:
(i) AC = AE (As given in the question)
(ii) ∠BAC = ∠EAD
(iii) AB = AD (It is also given in the question)
∴ Triangles ABC and ADE are similar i.e. ΔABC ≅ ΔADE.
So, by the rule of CPCT, it can be said that BC = DE.
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 (see Fig. 7.22). Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE
In the question, it is given that P is the mid-point of line segment AB. Also, ∠BAD = ∠ABE and ∠EPA = ∠DPB
(i) It is given that ∠EPA = ∠DPB
Now, add ∠DPE on both sides,
∠EPA +∠DPE = ∠DPB+∠DPE
This implies that angles DPA and EPB are equal i.e. ∠DPA = ∠EPB
Now, consider the triangles DAP and EBP.
∠DPA = ∠EPB
AP = BP (Since P is the mid-point of the line segment AB)
∠BAD = ∠ABE (As given in the question)
So, by ASA congruency, ΔDAP ≅ ΔEBP.
(ii) By the rule of CPCT, AD = BE.
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 (see Fig. 7.23). Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = ½ AB
It is given that M is the mid-point of the line segment AB, ∠C = 90°, and DM = CM
(i) Consider the triangles ΔAMC and ΔBMD:
AM = BM (Since M is the mid-point)
CM = DM (Given in the question)
∠CMA = ∠DMB (They are vertically opposite angles)
So, by SAS congruency criterion, ΔAMC ≅ ΔBMD.
(ii) ∠ACM = ∠BDM (by CPCT)
∴ AC || BD as alternate interior angles are equal.
Now, ∠ACB +∠DBC = 180° (Since they are co-interiors angles)
⇒ 90° +∠B = 180°
∴ ∠DBC = 90°
(iii) In ΔDBC and ΔACB,
BC = CB (Common side)
∠ACB = ∠DBC (They are right angles)
DB = AC (by CPCT)
So, ΔDBC ≅ ΔACB by SAS congruency.
(iv) DC = AB (Since ΔDBC ≅ ΔACB)
⇒ DM = CM = AM = BM (Since M the is mid-point)
So, DM + CM = BM+AM
Hence, CM + CM = AB
⇒ CM = (½) AB
Exercise 7.2
Question 1.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
AB = AC and
the bisectors of ∠B and ∠C intersect each other at O
(i) Since ABC is an isosceles with AB = AC,
∠B = ∠C
½ ∠B = ½ ∠C
⇒ ∠OBC = ∠OCB (Angle bisectors)
∴ OB = OC (Side opposite to the equal angles are equal.)
(ii) In ΔAOB and ΔAOC,
AB = AC (Given in the question)
AO = AO (Common arm)
OB = OC (As Proved Already)
So, ΔAOB ≅ ΔAOC by SSS congruence condition.
BAO = CAO (by CPCT)
Thus, AO bisects ∠A.
In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.
Since AD is bisector of BC.
∴BD = CD
Now, in ∆ABD and ∆ACD, we have
AD = DA [Common]
∠ADB = ∠ADC [Each 90°]
BD = CD [Proved above]
∴∆ABD ≅∆ACD [By SAS congruency]
⇒AB = AC [By C.P.C.T.]
Thus, ∆ABC is an isosceles triangle.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.
∆ABC is an isosceles triangle.
∴AB = AC
⇒∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal]
⇒∠BCE = ∠CBF
Now, in ∆BEC and ∆CFB
∠BCE = ∠CBF [Proved above]
∠BEC = ∠CFB [Each 90°]
BC = CB [Common]
∴∆BEC ≅∆CFB [By AAS congruency]
So, BE = CF [By C.P.C.T.]
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure).
Show that
(i) ∆ABE ≅∆ACF
(ii) AB = AC i.e., ABC is an isosceles triangle.
(i) In ∆ABE and ∆ACE, we have
∠AEB = ∠AFC
[Each 90° as BE ⊥AC and CF ⊥AB]
∠A = ∠A [Common]
BE = CF [Given]
∴∆ABE ≅∆ACF [By AAS congruency]
(ii) Since, ∆ABE ≅∆ACF
∴AB = AC [By C.P.C.T.]
⇒ABC is an isosceles triangle.
ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.
In ∆ABC, we have
AB = AC [ABC is an isosceles triangle]
∴∠ABC = ∠ACB …(1)
[Angles opposite to equal sides of a ∆ are equal]
Again, in ∆BDC, we have
BD = CD [BDC is an isosceles triangle]
∴∠CBD = ∠BCD …(2)
[Angles opposite to equal sides of a A are equal]
Adding (1) and (2), we have
∠ABC + ∠CBD = ∠ACB + ∠BCD
⇒∠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.
AB = AC [Given] …(1)
AB = AD [Given] …(2)
From (1) and (2), we have
AC = AD
Now, in ∆ABC, we have
∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]
⇒2∠ACB + ∠BAC = 180° …(3)
[∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)]
Similarly, in ∆ACD,
∠ADC + ∠ACD + ∠CAD = 180°
⇒2∠ACD + ∠CAD = 180° …(4)
[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°
⇒2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]
⇒2∠BCD = 360° – 180° = 180°
⇒∠BCD = 
Thus, ∠BCD = 90°
ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.
In ∆ABC, we have AB = AC [Given]
∴Their opposite angles are equal.
⇒∠ACB = ∠ABC
Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]
⇒90° + ∠B + ∠C = 180° [∠A = 90°(Given)]
⇒∠B + ∠C= 180°- 90° = 90°
But ∠B = ∠C
∠B = ∠C = 
Thus, ∠B = 45° and ∠C = 45°
Show that the angles of an equilateral triangle are 60° each.
In ∆ABC, we have
AB = BC = CA
[ABC is an equilateral triangle]
AB = BC
⇒∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]
Similarly, AC = BC
⇒∠A = ∠B …(2)
From (1) and (2), we have
∠A = ∠B = ∠C = x (say)
Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]
∴x + x + x = 180o
⇒3x = 180°
⇒x = 60°
∴∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.
Exercise 7.3
(i) 
(ii) 
(iii) AP bisects 
(iv) AP is the perpendicular bisector of BC.
(i)
Now in 
AB = AC [Given]
BD = CD [Given]
AD = AD [Common]
(ii)Now in 
AB = AC [Given]
AP = AP
(iii)Since 
Since 
Now 
And 
From eq. (iii) and (iv),
(iv)Since 
And 
Now 
From eq. (v), we have BP PC and from (vii), we have proved AP 
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC.
(ii) AD bisects 
In ΔADB and ΔADC,
AB = AC [Given]
AD = AD [Common]
Also 
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that
(i) ∆ABC ≅∆PQR
(ii) ∆ABM ≅∆PQN
In ∆ABC, AM is the median.
∴BM = 
In ∆PQR, PN is the median.
∴QN = 
And BC = QR [Given]
⇒
⇒BM = QN …(3) [From (1) and (2)]
(i) In ∆ABM and ∆PQN, we have
AB = PQ , [Given]
AM = PN [Given]
BM = QN [From (3)]
∴∆ABM ≅∆PQN [By SSS congruency]
(ii) Since ∆ABM ≅∆PQN
⇒∠B = ∠Q …(4) [By C.P.C.T.]
Now, in ∆ABC and ∆PQR, we have
∠B = ∠Q [From (4)]
AB = PQ [Given]
BC = QR [Given]
∴∆ABC ≅∆PQR [By SAS congruency]
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Since BE ⊥AC [Given]
∴BEC is a right triangle such that ∠BEC = 90°
Similarly, ∠CFB = 90°
Now, in right ∆BEC and ∆CFB, we have
BE = CF [Given]
BC = CB [Common hypotenuse]
∠BEC = ∠CFB [Each 90°]
∴∆BEC ≅∆CFB [By RHS congruency]
So, ∠BCE = ∠CBF [By C.P.C.T.]
or ∠BCA = ∠CBA
Now, in ∆ABC, ∠BCA = ∠CBA
⇒AB = AC [Sides opposite to equal angles of a ∆ are equal]
∴ABC is an isosceles triangle.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥BC to show that ∠B = ∠C.
We have, AP ⊥BC [Given]
∠APB = 90° and ∠APC = 90°
In ∆ABP and ∆ACP, we have
∠APB = ∠APC [Each 90°]
AB = AC [Given]
AP = AP [Common]
∴∆ABP ≅∆ACP [By RHS congruency]
So, ∠B = ∠C [By C.P.C.T.]
Exercise 7.4
Show that in a right angled triangle, the hypotenuse is the longest side.
Let us consider ∆ABC such that ∠B = 90°
∴∠A + ∠B + ∠C = 180°
⇒∠A + 90°-+ ∠C = 180°
⇒∠A + ∠C = 90°
⇒∠A + ∠C = ∠B
∴∠B > ∠A and ∠B > ∠C
⇒Side opposite to ∠B is longer than the side opposite to ∠A
i.e., AC > BC.
Similarly, AC > AB.
Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side.
In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
∠ABC + ∠PBC = 180° [Linear pair]
and ∠ACB + ∠QCB = 180° [Linear pair]
But ∠PBC < ∠QCB [Given] ⇒180° – ∠PBC > 180° – ∠QCB
⇒∠ABC > ∠ACB
The side opposite to ∠ABC > the side opposite to ∠ACB
⇒AC > AB.
In figure, ∠B <∠A and ∠C <∠D. Show that AD < BC.
Since ∠A > ∠B [Given]
∴OB > OA …(1)
[Side opposite to greater angle is longer]
Similarly, OC > OD …(2)
Adding (1) and (2), we have
OB + OC > OA + OD
⇒BC > AD
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B >∠D.
Let us join AC.
Now, in ∆ABC, AB < BC [∵AB is the smallest side of the quadrilateral ABCD] ⇒BC > AB
⇒∠BAC > ∠BCA …(1)
[Angle opposite to longer side of A is greater]
Again, in ∆ACD, CD > AD
[ CD is the longest side of the quadrilateral ABCD]
⇒∠CAD > ∠ACD …(2)
[Angle opposite to longer side of ∆ is greater]
Adding (1) and (2), we get
∠BAC + ∠CAD > ∠BCA + ∠ACD
⇒∠A > ∠C
Similarly, by joining BD, we have ∠B > ∠D.
In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ.
In ∆PQR, PS bisects ∠QPR [Given]
∴∠QPS = ∠RPS
and PR > PQ [Given]
⇒∠PQS > ∠PRS [Angle opposite to longer side of A is greater]
⇒∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS]
∵Exterior ∠PSR = [∠PQS + ∠QPS]
and exterior ∠PSQ = [∠PRS + ∠RPS]
[An exterior angle is equal to the sum of interior opposite angles]
Now, from (1), we have
∠PSR = ∠PSQ.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Let us consider the ∆PMN such that ∠M = 90°
Since, ∠M + ∠N+ ∠P = 180°
[Sum of angles of a triangle is 180°]
∵∠M = 90° [PM ⊥l]
So, ∠N + ∠P = ∠M
⇒∠N < ∠M
⇒PM < PN …(1)
Similarly, PM < PN1…(2)
and PM < PN2…(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular line segment is the shortest line segment drawn on a line from a point not on it.
Exercise 7.5
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Let us consider a ∆ABC.
Draw l, the perpendicular bisector of AB.
Draw m, the perpendicular bisector of BC.
Let the two perpendicular bisectors l and m meet at O.
O is the required point which is equidistant from A, B and C.
Note: If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Let us consider a ∆ABC.
Draw m, the bisector of ∠C.
Let the two bisectors l and m meet at O.
Thus, O is the required point which is equidistant from the sides of ∆ABC.
Note: If we draw OM ⊥BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle. Point O is called incentre of the triangle.
In a huge park, people are concentrated at three points (see figure)
A: where these 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 exist.
Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?
[Hint The parlour should be equidistant from A, B and C.]
Let us join A and B, and draw l, the perpendicular bisector of AB.
Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O.
The point O is the required point where the ice cream parlour be set up.
Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O.
Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
It is an activity.
We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii).
∴The Fig. (ii) has more triangles.
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Frequently Asked Questions
The NCERT solution for Class 9 Chapter 7: Triangles is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 9 Chapter 7: Triangles is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 9 Maths Chapter 7 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 9 Maths Chapter 7: Triangles as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 9 Maths Chapter 7 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.