### Class 9 Mathematics Chapter 7 – Triangles NCERT solutions

**Exercise: 7.1 (Page No: 118)**

**1. In quadrilateral ACBD, AC = AD and AB bisect A (see Fig. 7.16). Show that ΔABC ΔABD. What can you say about BC and BD?**

**Solution:**

It is given that AC and AD are equal i.e. AC = AD and the line segment AB bisects A.

We will have to now prove that the two triangles ABC and ABD are similar i.e. **ΔABC ΔABD**

**Proof:**

Consider the triangles ΔABC and ΔABD,

(i) AC = AD (It is given in the question)

(ii) AB = AB (Common)

(iii) CAB = DAB (Since AB is the bisector of angle A)

So, by **SAS congruency criterion**, ΔABC ΔABD.

For the 2nd part of the question, BC and BD are of equal lengths by the rule of C.P.C.T.

**2. ABCD is a quadrilateral in which AD = BC and DAB = CBA (see Fig. 7.17). Prove that**

(i) ΔABD ΔBAC

(ii) BD = AC

(iii) ABD = BAC.

**Solution:**

The given parameters from the questions are DAB = CBA and AD = BC.

(i) ΔABD and ΔBAC are similar by SAS congruency as

AB = BA (It is the common arm)

DAB = CBA and AD = BC (These are given in the question)

So, triangles ABD and BAC are similar i.e. ΔABD ΔBAC. (Hence proved).

(ii) It is now known that ΔABD ΔBAC so,

BD = AC (by the rule of CPCT).

(iii) Since ΔABD ΔBAC so,

Angles ABD = BAC (by the rule of CPCT).

**3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.**

**Solution:**

It is given that AD and BC are two equal perpendiculars to AB.

We will have to prove that **CD is the bisector of AB**

Now,

Triangles ΔAOD and ΔBOC are similar by AAS congruency since:

(i) A = B (They are perpendiculars)

(ii) AD = BC (As given in the question)

(iii) AOD = BOC (They are vertically opposite angles)

∴ ΔAOD ΔBOC.

So, AO = OB (by the rule of CPCT).

Thus, CD bisects AB (Hence proved).

**4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ΔCDA.**

**Solution:**

It is given that p q and l m

**To prove:**

Triangles ABC and CDA are similar i.e. ΔABC ΔCDA

**Proof**:

Consider the ΔABC and ΔCDA,

(i) BCA = DAC and BAC = DCA Since they are alternate interior angles

(ii) AC = CA as it is the common arm

So, by **ASA congruency criterion, **ΔABC ΔCDA.

**5. Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see Fig. 7.20). Show that:**

(i) ΔAPB ΔAQB

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

**Solution:**

It is given that the line “*l*” is the bisector of angle A and the line segments BP and BQ are perpendiculars drawn from *l*.

(i) ΔAPB and ΔAQB are similar by AAS congruency because:

P = Q (They are the two right angles)

AB = AB (It is the common arm)

BAP = BAQ (As line *l *is the bisector of angle A)

So, ΔAPB ΔAQB.

(ii) By the rule of CPCT, BP = BQ. So, it can be said the point B is equidistant from the arms of A.

**6. In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.**

**Solution:**

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.

**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 BAD = ABE and EPA = DPB (see Fig. 7.22). Show that**

**(i) ΔDAP ΔEBP**

**(ii) AD = BE**

**Solutions:**

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.

**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 Fig. 7.23). Show that:**

(i) ΔAMC ΔBMD

(ii) DBC is a right angle.

(iii) ΔDBC ΔACB

(iv) CM = ½ AB

**Solution:**

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 (Page No: 123)**

**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

**Solution:**

Given:

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.

**2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.**

**Solution:**

It is given that AD is the perpendicular bisector of BC

**To prove:**

AB = AC

**Proof:**

In ΔADB and ΔADC,

AD = AD (It is the Common arm)

ADB = ADC

BD = CD (Since AD is the perpendicular bisector)

So, ΔADB ΔADC by **SAS congruency criterion**.

Thus,

AB = AC (by CPCT)

**3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.**

**Solution:**

Given:

(i) BE and CF are altitudes.

(ii) AC = AB

**To prove:**

BE = CF

**Proof:**

Triangles ΔAEB and ΔAFC are similar by AAS congruency since

A = A (It is the common arm)

AEB = AFC (They are right angles)

AB = AC (Given in the question)

∴ ΔAEB ΔAFC and so, BE = CF (by CPCT).

**4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that**

(i) ΔABE ΔACF

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

**Solution:**

It is given that BE = CF

(i) In ΔABE and ΔACF,

A = A (It is the common angle)

AEB = AFC (They are right angles)

BE = CF (Given in the question)

∴ ΔABE ΔACF by **AAS congruency condition**.

(ii) AB = AC by CPCT and so, ABC is an isosceles triangle.

**5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ABD = ACD.**

**Solution:**

In the question, it is given that ABC and DBC are two isosceles triangles.

We will have to show that ABD = ACD

**Proof:**

Triangles ΔABD and ΔACD are similar by SSS congruency since

AD = AD (It is the common arm)

AB = AC (Since ABC is an isosceles triangle)

BD = CD (Since BCD is an isosceles triangle)

So, ΔABD ΔACD.

∴ ABD = ACD by CPCT.

**6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that BCD is a right angle.**

**Solution:**

It is given that AB = AC and AD = AB

We will have to now prove BCD is a right angle.

**Proof:**

Consider ΔABC,

AB = AC (It is given in the question)

Also, ACB = ABC (They are angles opposite to the equal sides and so, they are equal)

Now, consider ΔACD,

AD = AB

Also, ADC = ACD (They are angles opposite to the equal sides and so, they are equal)

Now,

In ΔABC,

CAB + ACB + ABC = 180°

So, CAB + 2ACB = 180°

⇒ CAB = 180° – 2ACB — (i)

Similarly, in ΔADC,

CAD = 180° – 2ACD — (ii)

also,

CAB + CAD = 180° (BD is a straight line.)

Adding (i) and (ii) we get,

CAB + CAD = 180° – 2ACB+180° – 2ACD

⇒ 180° = 360° – 2ACB-2ACD

⇒ 2(ACB+ACD) = 180°

⇒ BCD = 90°

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

**Solution:**

In the question, it is given that

A = 90° and AB = AC

AB = AC

⇒ B = C (They are angles opposite to the equal sides and so, they are equal)

Now,

A+B+C = 180° (Since the sum of the interior angles of the triangle)

∴ 90° + 2B = 180°

⇒ 2B = 90°

⇒ B = 45°

So, B = C = 45°

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

**Solution:**

Let ABC be an equilateral triangle as shown below:

Here, BC = AC = AB (Since the length of all sides is same)

⇒ A = B =C (Sides opposite to the equal angles are equal.)

Also, we know that

A+B+C = 180°

⇒ 3A = 180°

⇒ A = 60°

∴ A = B = C = 60°

So, the angles of an equilateral triangle are always 60° each.

**Exercise: 7.3 (Page No: 128)**

**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 Fig. 7.39). If AD is extended to intersect BC at P, show that**

(i) ΔABD ΔACD

(ii) ΔABP ΔACP

(iii) AP bisects A as well as D.

(iv) AP is the perpendicular bisector of BC.

**Solution:**

In the above question, it is given that ΔABC and ΔDBC are two isosceles triangles.

(i) ΔABD and ΔACD are similar by SSS congruency because:

AD = AD (It is the common arm)

AB = AC (Since ΔABC is isosceles)

BD = CD (Since ΔDBC is isosceles)

∴ ΔABD ΔACD.

(ii) ΔABP and ΔACP are similar as:

AP = AP (It is the common side)

PAB = PAC (by CPCT since ΔABD ΔACD)

AB = AC (Since ΔABC is isosceles)

So, ΔABP ΔACP by SAS congruency condition.

(iii) PAB = PAC by CPCT as ΔABD ΔACD.

AP bisects A. — (i)

Also, ΔBPD and ΔCPD are similar by SSS congruency as

PD = PD (It is the common side)

BD = CD (Since ΔDBC is isosceles.)

BP = CP (by CPCT as ΔABP ΔACP)

So, ΔBPD ΔCPD.

Thus, BDP = CDP by CPCT. — (ii)

Now by comparing (i) and (ii) it can be said that AP bisects A as well as D.

(iv) BPD = CPD (by CPCT as ΔBPD ΔCPD)

and BP = CP — (i)

also,

BPD +CPD = 180° (Since BC is a straight line.)

⇒ 2BPD = 180°

⇒ BPD = 90° —(ii)

Now, from equations (i) and (ii), it can be said that

AP is the perpendicular bisector of BC.

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

(i) AD bisects BC (ii) AD bisects A.

**Solution:**

It is given that AD is an altitude and AB = AC. The diagram is as follows:

(i) In ΔABD and ΔACD,

ADB = ADC = 90°

AB = AC (It is given in the question)

AD = AD (Common arm)

∴ ΔABD ΔACD by RHS congruence condition.

Now, by the rule of CPCT,

BD = CD.

So, AD bisects BC

(ii) Again, by the rule of CPCT, BAD = CAD

Hence, AD bisects A.

**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 Fig. 7.40). Show that:**

(i) ΔABM ΔPQN

(ii) ΔABC ΔPQR

**Solution:**

Given parameters are:

AB = PQ,

BC = QR and

AM = PN

(i) ½ BC = BM and ½ QR = QN (Since AM and PN are medians)

Also, BC = QR

So, ½ BC = ½ QR

⇒ BM = QN

In ΔABM and ΔPQN,

AM = PN and AB = PQ (As given in the question)

BM = QN (Already proved)

∴ ΔABM ΔPQN by SSS congruency.

(ii) In ΔABC and ΔPQR,

AB = PQ and BC = QR (As given in the question)

ABC = PQR (by CPCT)

So, ΔABC ΔPQR by SAS congruency.

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

.

**Solution:**

It is known that BE and CF are two equal altitudes.

Now, in ΔBEC and ΔCFB,

BEC = CFB = 90° (Same Altitudes)

BC = CB (Common side)

BE = CF (Common side)

So, ΔBEC ΔCFB by RHS congruence criterion.

Also, C = B (by CPCT)

Therefore, AB = AC as sides opposite to the equal angles is always equal.

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

**Solution:**

In the question, it is given that AB = AC

Now, ΔABP and ΔACP are similar by RHS congruency as

APB = APC = 90° (AP is altitude)

AB = AC (Given in the question)

AP = AP (Common side)

So, ΔABP ΔACP.

∴ B = C (by CPCT)

**Exercise: 7.4 (Page No: 132)**

**1. Show that in a right-angled triangle, the hypotenuse is the longest side.**

**Solution:**

It is known that ABC is a triangle right angled at B.

We know that,

A +B+C = 180°

Now, if B+C = 90° then A has to be 90°.

Since A is the largest angle of the triangle, the side opposite to it must be the largest.

So, AB is the hypotenuse which will be the largest side of the above right-angled triangle i.e. ΔABC.

**2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.**

**Solution:**

It is given that PBC < QCB

We know that ABC + PBC = 180°

So, ABC = 180°-PBC

Also,

ACB +QCB = 180°

Therefore ACB = 180° -QCB

Now, since PBC < QCB,

∴ ABC > ACB

Hence, AC > AB as sides opposite to the larger angle is always larger.

**3. In Fig. 7.49, B < A and C < D. Show that AD < BC.**

**Solution:**

In the question, it is mentioned that angles B and angle C is smaller than angles A and D respectively i.e. B < A and C < D.

Now,

Since the side opposite to the smaller angle is always smaller

AO < BO — (i)

And OD < OC —(ii)

By adding equation (i) and equation (ii) we get

AO+OD < BO + OC

So, AD < BC

**4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).**

**Show that A > C and B > D.**

**Solution:**

In ΔABD, we see that

AB < AD < BD

So, ADB < ABD — (i) (Since angle opposite to longer side is always larger)

Now, in ΔBCD,

BC < DC < BD

Hence, it can be concluded that

BDC < CBD — (ii)

Now, by adding equation (i) and equation (ii) we get,

ADB + BDC < ABD + CBD

ADC < ABC

B > D

Similarly, In triangle ABC,

ACB < BAC — (iii) (Since the angle opposite to the longer side is always larger)

Now, In ΔADC,

DCA < DAC — (iv)

By adding equation (iii) and equation (iv) we get,

ACB + DCA < BAC+DAC

⇒ BCD < BAD

∴ A > C

**5. In Fig 7.51, PR > PQ and PS bisect QPR. Prove that PSR > PSQ.**

**Solution:**

It is given that PR > PQ and PS bisects QPR

Now we will have to prove that angle PSR is smaller than PSQ i.e. PSR > PSQ

**Proof:**

QPS = RPS — (ii) (As PS bisects ∠QPR)

PQR > PRQ — (i) (Since PR > PQ as angle opposite to the larger side is always larger)

PSR = PQR + QPS — (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)

PSQ = PRQ + RPS — (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)

By adding (i) and (ii)

PQR +QPS > PRQ +RPS

Thus, from (i), (ii), (iii) and (iv), we get

PSR > PSQ

**6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.**

**Solution:**

First, let “*l*” be a line segment and “B” be a point lying on it. A line AB perpendicular to *l* is now drawn. Also, let C be any other point on *l*. The diagram will be as follows:

**To prove:**

AB < AC

**Proof:**

In ΔABC, B = 90°

Now, we know that

A+B+C = 180°

∴ A +C = 90°

Hence, C must be an acute angle which implies C < B

So, AB < AC (As the side opposite to the larger angle is always larger)