NCERT Solutions for Class 9 Maths Chapter 7 - Triangles

1. In quadrilateral ACBD, AC = AD and AB bisects \[\angle A\] (See the given figure). Show that \[\Delta ABC{\text{ }} \cong \Delta ABD\]. What can you say about BC and BD?

Quadrilateral ABCD

Ans: Given: In quadrilateral ACBD, AC = AD and AB is bisected by \[\angle A\]

To find: Show that \[\Delta ABC{\text{ }} \cong \Delta ABD\]. 

In \[\Delta ABC{\text{ , }}\Delta ABD\]

\[AC{\text{ }} = {\text{ }}AD\] (Given)

\[\angle CAB{\text{ }} = \angle DAB\]     (AB bisects ∠A)

\[AB{\text{ }} = {\text{ }}AB\]        (Common)

\[\therefore \Delta ABC{\text{ }} \cong \Delta ABD\]            (By SAS congruence rule)

\[\therefore BC{\text{ }} = {\text{ }}BD\]     (By CPCT)

Therefore, BC and BD are of equal lengths.

2. ABCD is a quadrilateral in which AD = BC and \[\angle DAB{\text{ }} = \angle CBA\]  (See the given figure). Prove that

(i)  \[\Delta ABD{\text{ }} \cong \Delta BAC\]

(ii) BD = AC

(iii)\[\angle ABD{\text{ }} = \angle BAC\].

Quadrilateral ABCD with AD = BC

Ans : Given: ABCD is a quadrilateral where AD = BC and \[\angle DAB{\text{ }} = \angle CBA\]

To prove: 

(i)  \[\Delta ABD{\text{ }} \cong \Delta BAC\]

(ii) BD = AC

(iii)\[\angle ABD{\text{ }} = \angle BAC\].

In \[\Delta ABD{\text{ , }}\Delta BAC\],

AD = BC    (Given)

\[\angle DAB{\text{ }} = \angle CBA\]   (Given)

AB = BA     (Common)

\[\therefore \Delta ABD{\text{ }} \cong \Delta BAC\]              (By SAS congruence rule)

\[\therefore BD{\text{ }} = {\text{ }}AC\]          (By CPCT)

And, \[\angle ABD{\text{ }} = \angle BAC\]           (By CPCT)

3. AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.

Figure showing AD and BC equal perpendiculars to a line segment AB

Ans: Given: AD and BC are equal perpendiculars to a line segment AB 

To prove: CD bisects AB.

In \[\Delta BOC{\text{ , }}\Delta AOD\],

\[\angle BOC{\text{ }} = \angle AOD\]     (Vertically opposite angles)

\[\angle CBO{\text{ }} = \angle DAO\]        (Each right angle )

BC = AD     (Given)

\[\therefore \Delta BOC{\text{ }} \cong \Delta AOD\]     (AAS congruence rule)

\[\therefore BO{\text{ }} = {\text{ }}AO\]     (By CPCT)

\[ \Rightarrow \]CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that \[\Delta ABC \cong \Delta CDA\].

Two  Parallel Lines L and M

Given: l and m are two parallel lines intersected by another pair of parallel lines p and q To prove:  \[\Delta ABC \cong \Delta CDA\].

In \[\Delta ABC{\text{ , }}\Delta CDA\],

\[\angle BAC{\text{ }} = \angle DCA\]       (Alternate interior angles, as\[p{\text{ }}||{\text{ }}q\])

2AC = CA     (Common)

\[\angle BCA{\text{ }} = \angle DAC\]       (Alternate interior angles, as \[l{\text{ }}||{\text{ }}m\])

\[\therefore \Delta ABC \cong \Delta CDA\]          (By ASA congruence rule)

5. Line l is the bisector of an angle \[\angle A\] and B is any point on l. BP and BQ are $$perpendiculars from B to the arms of \[\angle A\] (see the given figure). Show that:

Line l the Bisector of an Angle A

(i) \[\Delta APB \cong \Delta AQB\]

(ii) BP = BQ or B is equidistant from the arms of ∠A.

Ans: Given: Line l is the bisector of an angle \[\angle A\] and B is any point on l.

To prove: (i) \[\Delta APB \cong \Delta AQB\]

(ii) \[BP{\text{ }} = {\text{ }}BQ\] or B is equidistant from the arms of \[\angle A\].

In \[\Delta APB{\text{ , }}\Delta AQB\],

\[\angle APB{\text{ }} = \angle AQB\]        (Each right angle )

\[\angle PAB{\text{ }} = \angle QAB\]      (l is the angle bisector of \[\angle A\])

AB = AB (Common)

\[\therefore \Delta APB \cong \Delta AQB\]       (By AAS congruence rule)

\[\therefore BP{\text{ }} = {\text{ }}BQ\]        (By CPCT)

Or, it can be said that B is equidistant from the arms of \[\angle A\].

6. In the given figure, \[AC{\text{ }} = {\text{ }}AE,{\text{ }}AB{\text{ }} = {\text{ }}AD\] and \[\angle BAD{\text{ }} = \angle EAC\]. Show that BC = DE.

Ans:

Angle BAD = angle EAC

Given: \[\angle BAD{\text{ }} = \angle EAC\]

To prove: BC = DE

It is given that \[\angle BAD{\text{ }} = \angle EAC\]

\[\angle BAD{\text{ }} + \angle DAC{\text{ }} = \angle EAC{\text{ }} + \angle DAC\]

\[\angle BAC{\text{ }} = \angle DAE\]

In \[\Delta BAC{\text{ , }}\Delta DAE\],

AB = AD     (Given)

\[\angle BAC{\text{ }} = \angle DAE\]      (Proved above)

AC = AE    (Given)

\[\therefore \Delta BAC{\text{ }} \cong \Delta DAE\]       (By SAS congruence rule)

\[\therefore BC{\text{ }} = {\text{ }}DE\]           (By CPCT)

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \[\angle BAD{\text{ }} = \angle ABE\] and \[\angle EPA{\text{ }} = \angle DPB\] (See the given figure). Show that

Line segment AB with its mid point P

(i) \[\Delta DAP \cong \Delta EBP\]

(ii) AD = BE

Ans: Given: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \[\angle BAD{\text{ }} = \angle ABE\] and \[\angle EPA{\text{ }} = \angle DPB\]

To prove: (i) \[\Delta DAP \cong \Delta EBP\]

(ii) AD = BE

It is given that \[\angle EPA{\text{ }} = \angle DPB\]

\[\angle EPA{\text{ }} + \angle DPE{\text{ }} = \angle DPB{\text{ }} + \angle DPE\]

\[\therefore \angle DPA{\text{ }} = \angle EPB\]

In \[\Delta DAP{\text{ , }}\Delta EBP\],

\[\angle DAP{\text{ }} = \angle EBP\] (Given)

AP = BP        (P is mid-point of AB)

\[\angle DPA{\text{ }} = \angle EPB\]   (From above)

\[\therefore \Delta DAP \cong \Delta EBP\]    (ASA congruence rule)

\[\therefore AD{\text{ }} = {\text{ }}BE\]          (By CPCT)

8. 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 the given figure). Show that:

Triangle ABC right - angled at C and midpoint M of AB

i) \[\Delta AMC \cong \Delta BMD\]

ii) \[\angle DBC\] is a right angle.

iii) \[\Delta DBC \cong \Delta ACB\]

iv) \[CM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\]

Ans: Given: M is the mid-point of hypotenuse AB. DM = CM

(i) In \[\Delta AMC{\text{ }},{\text{ }}\Delta BMD\],

AM = BM (M is the mid-point of AB)

\[\angle AMC{\text{ }} = \angle BMD\]        (Vertically opposite angles)

CM = DM          (Given)

\[\therefore \Delta AMC \cong \Delta BMD\]          (By SAS congruence rule)

\[\therefore AC{\text{ }} = {\text{ }}BD\] (By CPCT)

And, \[\angle ACM{\text{ }} = \angle BDM\](By CPCT)

(ii) \[\angle ACM{\text{ }} = \angle BDM\]

However,  \[\angle ACM{\text{ , }}\angle BDM\] are alternate interior angles.

Since alternate angles are equal,

It can be said that \[DB{\text{ }}||{\text{ }}AC\]

(Co-interior angles)

(iii) In \[\Delta DBC{\text{ , }}\Delta ACB\],

DB = AC      (Already proved)

\[\angle DBC{\text{ }} = \angle ACB\]          (Each \[{90^o}\])

BC = CB (Common)

\[\therefore \Delta DBC \cong \Delta ACB\]           (SAS congruence rule)

(iv) \[\therefore \Delta DBC \cong \Delta ACB\]

AB = DC (By CPCT)

AB = 2 CM

\[CM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\]

Exercise- 7.2

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

Isosceles triangle with AB = AC

Ans: We know, $\angle ABC = \angle ACB$ ${\text{[Equal angles of isosceles triangle]}}$

$ \Rightarrow \dfrac{1}{2}\angle ABC = \dfrac{1}{2}ACB$

$ \Rightarrow \angle OBC = \angle OCB$

$\therefore OB = OC$ ${\text{[Sides opposite to equal angles of Isosceles triangle are equal]}}$

Now, In $\Delta ABO$ and $\Delta ACO$,

$AB = AC$ ${\text{[Equal sides of Isosceles triangle]}}$

$OB = OC$ ${\text{[Proved above]}}$

AO=AO [Common]

$\therefore \Delta ABO \cong \Delta ACO$ ${\text{[By SSS]}}$

$\therefore \angle BAO = \angle CAO$ ${\text{[CPCT]}}$

$\therefore $ AO bisects angle A.

2. In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.

Triangle ABC , AD being the bisector of BC

Ans: $\because $ AD is perpendicular bisector of BC,

$\therefore $$BD = DC$

& $\angle ADB = \angle ADC$ ${\text{[Each 90°]}}$

Now, in$\Delta ABD$ and $\Delta ACD$,

$AD = AD$ (Common)

$\angle ADB = \angle ADC$ ${\text{[Proved above]}}$ $BD = CD$ ${\text{[Proved above]}}$

$\therefore \Delta ADB \cong \Delta ACD$ ${\text{[by SAS]}}$

$\therefore AB = AC$ (CPCT)

So, $\Delta ABC$ is an isosceles triangle with \[AB = AC\]

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

Isosceles triangle ABC

Ans: In $\Delta ABE$and $\Delta ACF$,

$\angle AEB = \angle AFE$ ${\text{[Each 90°]}}$

$\angle BAE = \angle CAF$ (Common)

$AB = AC$ [Given]

$\therefore \Delta ABE \cong \Delta ACF$ ${\text{[By AAS]}}$

$\therefore BE = CF$ (CPCT)

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that

(i) \[\Delta ABE \cong \Delta ACF\]

(ii) AB = AC i.e., ABC is an isosceles triangle

Triangle ABC with equal altitudes

Ans: In $\Delta ABE$and $\Delta ACF$,

\[\angle BAE = \angle CAF\] (Common)

$\angle BEA = \angle CFA$               ${\text{[Each 90°]}}$

$BE = CF$ [Given]

$\therefore \Delta ABE \cong \Delta ACF$ ${\text{[by AAS]}}$

$\therefore AB = AC$ (CPCT)

And therefore, $\Delta ABC$ is an isosceles triangle with AB=AC.

5. ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.

Two Isosceles Triangles

Ans: Since AB=AC, and DB=DC,

$\therefore $ABDC is a quadrilateral with adjacent sides being equal.

$ \Rightarrow $ABDC is a kite.

We know that one pair of opposite (obtuse) angles of kite are equal.

Hence, $\angle ABD = \angle ACD$

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.

Isosceles Triangle in which AB = AC

Ans: In $\Delta ABC$

$\angle ABC = \angle ACB$ ${\text{[angles opposite to equal sides of a triangle]}}$ … (1)

Similarly, in $\Delta ADC$

$\angle ADC = \angle ACD$ ${\text{[angles opposite to equal sides of a triangle]}}$ … (2)

Now since BD is a straight line,

$\angle BAC + \angle CAD = 180^\circ $ … (3)

And we know,

$\angle BAC = \angle ADC + \angle ACD$ ${\text{[Exterior angle of a triangle = sum of opposite interior angles]}}$

$ \Rightarrow \angle BAC = 2\angle ACD$ ${\text{[From 2]}}$ … (4)

Similarly,

$\angle CAD = \angle ABC + \angle ACB$

$ \Rightarrow \angle CAD = 2\angle ACB$ ${\text{[From 1]}}$ … (5)

$\therefore 2\angle ACB + 2\angle ACD = 180^\circ $         ${\text{[From 3, 4 and 5]}}$

$ \Rightarrow \angle ACB + \angle ACD = 90^\circ  $

$ \Rightarrow \angle BCD = 90^\circ  $

Hence proved.

7. ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Right - Angled Triangle ABC

Ans: Since AB = AC,

$\angle B = \angle C$ ${\text{[Angles opposite to equal sides of a triangle]}}$

Now, we know

$\angle A + \angle B + \angle C = 180^\circ $ ${\text{[Angle sum property]}}$

$ \Rightarrow \angle A + 2\angle C = 180^\circ $

$ \Rightarrow 90^\circ  + 2\angle C = 180^\circ  $

$ \Rightarrow 2\angle C = 90^\circ  $

$  \Rightarrow \angle C = 45^\circ   $

$\therefore \angle B = 45^\circ $

Hence $\angle B$ and $\angle C$ are each 45$^\circ $

8. Show that the angles of an equilateral triangle are 60º each.

Ans:

Equilateral Triangle

In $\angle B = \angle C$ ${\text{[Angles opposite to equal sides of a triangle]}}$

Similarly,

$\angle A = \angle B$ ${\text{[Angles opposite to equal sides of a triangle]}}$

$ \Rightarrow \angle A = \angle B = \angle C$

Now, we know that

$\angle A + \angle B + \angle C = 180^\circ $ ${\text{[Angle sum property]}}$

$ \Rightarrow \angle A + \angle A + \angle A = 180^\circ  $

$\Rightarrow 3\angle A = 180^\circ  $

$ \Rightarrow \angle A = 60^\circ $ 

$\therefore \angle A = \angle B = \angle C = 60^\circ $

Hence proved.

Exercise- 7.3

1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that

(i) \[\Delta ABD \cong \Delta ACD\]

(ii) \[\Delta ABP \cong \Delta ACP\]

(iii) AP bisects $\angle A$ as well as $\angle D$

(iv)AP is the bisector of BC

Two Isosceles Triangle on the same Base

Ans: (i) In $\Delta ABD$ and $\Delta ACD$,

$AB = AC$ ${\text{[Equal sides of isosceles triangle]}}$

$DB = DC$ ${\text{[Equal sides of isosceles triangle]}}$

AD=AD (Common)

$\therefore \Delta ABD \cong \Delta ACD$ ${\text{[By SSS]}}$

$ \Rightarrow \angle BAD = \angle CAD$ [CPCT]

$ \Rightarrow \angle BAP = \angle CAP$ … (1)

And $\angle ADB = \angle ADC$ (CPCT) … (2)

(ii) In $\Delta ABP$ and $\Delta ACP$

$AB = AC$ ${\text{[Equal sides of isosceles triangle]}}$

$\angle BAP = \angle CAP$ ${\text{[From 1]}}$

$AP = AP$ (Common)

$\therefore \Delta ABP \cong \Delta ACP$ ${\text{[By SAS]}}$

$\therefore BP = CP$ (CPCT) … (3)

Similarly, $\angle APB = \angle APC$ (CPCT) … (4)

(iii) AP is bisector of $\angle A$ ${\text{[From 1]}}$

Now, since AP is a line segment

$\therefore \angle ADB + \angle BDP = 180^\circ $ … (5)

Similarly, $\angle ADC + \angle CDP = 180^\circ $ … (6)

Comparing equations 2, 5 and 6 we can say that

$\angle BDP = \angle CDP$

$\therefore $AP bisects $\angle D$

Hence AP bisects both $\angle A$ and $\angle D$

(iv)We know,

$\angle APB + \angle APC = 180^\circ $

$ \Rightarrow \angle APB + \angle APB = 180^\circ $ ${\text{[From 4]}}$

$ \Rightarrow \angle APB = 90^\circ $ … (7)

From equations 3 and 7 we can say that,

AP is perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects $\angle A$

Ans:

Isosceles Triangle BC with AB = AC

In $\Delta ADB$ and $\Delta ADC$

$AB = AC$ (Given)

$AD = AD$ (Common)

$\angle ADB = \angle ADC$ ${\text{[Each 90°]}}$

$\therefore \Delta ADB \cong \Delta ADC$ ${\text{[By RHS]}}$

$ \Rightarrow BD = DC$ (CPCT)

Therefore, AD is bisector of BC

Similarly, $\angle BAD = \angle CAD$ (CPCT)

Therefore, AD bisects $\angle A$as well.

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:

(i)\[\Delta ABM \cong \Delta PQN\]

(ii)\[\Delta ABC \cong \Delta PQR\]

Two Triangles with Their Corresponding Parts

Ans: (i) We know that $BC = QR$ … (1)

Now, Since AM is median of ΔABC,

$ \Rightarrow BM = \dfrac{1}{2}BC$ … (2)

Similarly, PN is median of ΔPQR,

$ \Rightarrow QN = \dfrac{1}{2}QR$ … (3)

From equations 1, 2 and 3, we can say that,

$BM = QN$ … (4)

Now in $\Delta ABM$ and $\Delta PQN$

\[AB = PQ\] (Given)

\[BM = QN\] ${\text{[From 4]}}$

AM=PN (Given)

$\therefore \Delta ABM \cong \Delta PQN$ ${\text{[By SSS]}}$

$ \Rightarrow \angle ABM = \angle AQN$ (CPCT)

$ \Rightarrow \angle ABC = \angle PQR$ … (5)

(ii) Now in $\Delta ABC$ and $\Delta PQR$,

$AB = PQ$ (Given)

$\angle ABC = \angle PQR$ ${\text{[From 5]}}$

$BC = QR$ (Given)

\[\Delta ABC \cong \Delta PQR\]         ${\text{[By SAS]}}$

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Ans:

Two Equal Altitudes BE and CF of Triangle ABC

In $\Delta BEC$ and $\Delta CFB$

$BE = CF$ (Given)

$\angle BEC = \angle CFB$ ${\text{[Each 90°]}}$

BC=CB (Common)

$\therefore \Delta BEC \cong \Delta CFB$ ${\text{[By RHS Congruency]}}$

$ \Rightarrow \angle BCE = \angle CBF$               (CPCT)

$\therefore AB = AC$ ${\text{[Sides opposite to equal angles of a triangle are equal]}}$

Therefore, ΔABC is an isosceles triangle.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B= ∠C.

Isosceles Triangle ABC with AB = AC

Ans: In $\Delta ABP$ and $\Delta ACP$

$AB = AC$ (Given)

$AP = AP$ (Common)

$\angle APB = \angle APC$ ${\text{[Each 90°]}}$

$\therefore \Delta ABP \cong \Delta ACP$ ${\text{[By RHS]}}$

$ \Rightarrow \angle ABP = \angle ACP$         (CPCT)

Overview of Deleted Syllabus for CBSE Class 9 Maths Chapter 7 Triangles

Class 9 Maths Chapter 7: Exercises Breakdown

Conclusion

Triangle-related NCERT Solutions for Class 9 Chapter 7 from Vedantu are an excellent resource for understanding important geometric ideas. Because it provides the foundation for understanding more difficult geometrical structures that students will encounter in higher grades, this chapter is essential. Along with several congruence criteria including SAS, SSS, and ASA, it concentrates on key theorems such as the Pythagorean Theorem, Angle Sum Property, and Triangle Inequality. The importance of understanding both the theoretical and practical elements of these geometric principles is made clear by the three to five problems from this chapter that have frequently appeared in earlier year's question papers.

Other Study Material for CBSE Class 9 Maths Chapter 7

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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