#### NCERT Solutions for Class 9th: Ch 7 Triangles Maths

Page No: 118

Exercise 7.1

1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that Î”ABC ≅ Î”ABD. What can you say about BC and BD?

Given,
AC = AD and AB bisects ∠A
To prove,
Î”ABC ≅ Î”ABD
Proof,
In Î”ABC and Î”ABD,
AB = AB (Common)
∠CAB = ∠DAB (AB is bisector)
Therefore, Î”ABC ≅ Î”ABD by SAS congruence condition.
BC and BD are of equal length.

Page No: 119

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that
(i) Î”ABD ≅ Î”BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.

Given,
AD = BC and ∠DAB = ∠CBA

(i) In Î”ABD and Î”BAC,
AB = BA (Common)
∠DAB = ∠CBA (Given)
Therefore, Î”ABD ≅ Î”BAC by SAS congruence condition.
(ii) Since, Î”ABD ≅ Î”BAC
Therefore BD = AC by CPCT
(iii) Since, Î”ABD ≅ Î”BAC
Therefore ∠ABD = ∠BAC by CPCT

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

Given,
AD and BC are equal perpendiculars to AB.
To prove,
CD bisects AB
Proof,
In Î”AOD and Î”BOC,
∠A = ∠B (Perpendicular)
∠AOD = ∠BOC (Vertically opposite angles)
Therefore, Î”AOD ≅ Î”BOC by AAS congruence condition.
Now,
AO = OB (CPCT). CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that Î”ABC ≅ Î”CDA.

Given,
l || m and p || q
To prove,
Î”ABC ≅ Î”CDA
Proof,
In Î”ABC and Î”CDA,
∠BCA = ∠DAC (Alternate interior angles)
AC = CA (Common)
∠BAC = ∠DCA (Alternate interior angles)
Therefore, Î”ABC ≅ Î”CDA by ASA congruence condition.

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) Î”APB ≅ Î”AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

Given,
l is the bisector of an angle ∠A.
BP and BQ are perpendiculars.

(i) In Î”APB and Î”AQB,
∠P = ∠Q (Right angles)
∠BAP = ∠BAQ (l is bisector)
AB = AB (Common)
Therefore, Î”APB ≅ Î”AQB by AAS congruence condition.
(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of ∠A.

Page No: 120

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

Given,
To show,
BC = DE
Proof,
∠BAD + ∠DAC = ∠EAC + ∠DAC
AC = AE (Given)
Therefore, Î”ABC ≅ Î”ADE by SAS congruence condition.
BC = 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 ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that
(i) Î”DAP ≅ Î”EBP

Given,
P is mid-point of AB.
∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)
∠EPA + ∠DPE = ∠DPB + ∠DPE
⇒ ∠DPA = ∠EPB
In Î”DAP ≅ Î”EBP,
∠DPA = ∠EPB
AP = BP (P is mid-point of AB)
Therefore, Î”DAP ≅ Î”EBP by ASA congruence condition.
(ii) AD = 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 Fig. 7.23). Show that:
(i) Î”AMC ≅ Î”BMD
(ii) ∠DBC is a right angle.
(iii) Î”DBC ≅ Î”ACB
(iv) CM = 1/2 AB

Given,
∠C = 90°, M is the mid-point of AB and DM = CM

(i) In Î”AMC and Î”BMD,
AM = BM (M is the mid-point)
∠CMA = ∠DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, Î”AMC ≅ Î”BMD by SAS congruence condition.

(ii) ∠ACM = ∠BDM (by CPCT)
Therefore, AC || BD as alternate interior angles are equal.
Now,
∠ACB + ∠DBC = 180° (co-interiors angles)
⇒ 90° + ∠B = 180°
⇒ ∠DBC = 90°

(iii) In Î”DBC and Î”ACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC (byy CPCT, already proved)
Therefore, Î”DBC ≅ Î”ACB by SAS congruence condition.

(iv)  DC = AB (Î”DBC ≅ Î”ACB)
⇒ DM = CM = AM = BM (M is mid-point)
⇒ DM + CM = AM + BM
⇒ CM + CM = AB
⇒ CM = 1/2AB

Page No: 123

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

Given,
AB = AC, the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,
∴ ∠B = ∠C
⇒ 1/2∠B = 1/2∠C
⇒ ∠OBC = ∠OCB (Angle bisectors.)
⇒ OB = OC (Side opposite to the equal angles are equal.)

(ii) In Î”AOB and Î”AOC,
AB = AC (Given)
AO = AO (Common)
OB = OC (Proved above)
Therefore, Î”AOB ≅ Î”AOC by SSS congruence condition.
∠BAO = ∠CAO (by CPCT)
Thus, AO bisects ∠A.

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

Given,
AD is the perpendicular bisector of BC
To show,
AB = AC
Proof,
BD = CD (AD is the perpendicular bisector)
AB = AC (by CPCT)

Page No: 124

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.

Given,
BE and CF are altitudes.
AC = AB
To show,
BE = CF
Proof,
In Î”AEB and Î”AFC,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
AB = AC (Given)
Therefore, Î”AEB ≅ Î”AFC by AAS congruence condition.
Thus, 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) Î”ABE ≅ Î”ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Given,
BE = CF

(i) In Î”ABE and Î”ACF,
∠A = ∠A (Common)
∠AEB = ∠AFC (Right angles)
BE = CF (Given)
Therefore, Î”ABE ≅ Î”ACF by AAS congruence condition.

(ii) Thus, AB = AC by CPCT and therefore 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.

Given,
ABC and DBC are two isosceles triangles.
To show,
∠ABD = ∠ACD
Proof,
In Î”ABD and Î”ACD,
AB = AC (ABC is an isosceles triangle.)
BD = CD (BCD is an isosceles triangle.)
Therefore, Î”ABD ≅ Î”ACD by SSS congruence condition. Thus, ∠ABD = ∠ACD by CPCT.

6. Î”ABC 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.

Given,
AB = AC and AD = AB
To show,
∠BCD is a right angle.
Proof,
In Î”ABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)
In Î”ACD,
⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)
Now,
In Î”ABC,
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° - 2∠ACB --- (i)
∠CAD = 180° - 2∠ACD --- (ii)
also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
∠CAB + ∠CAD = 180° - 2∠ACB + 180° - 2∠ACD
⇒ 180° = 360° - 2∠ACB - 2∠ACD
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠BCD = 90°

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

Given,
∠A = 90° and AB = AC
A/q,
AB = AC
⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)
Now,
∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)
⇒ 90° + 2∠B = 180°
⇒ 2∠B = 90°
⇒ ∠B = 45°
Thus, ∠B = ∠C = 45°

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

Let ABC be an equilateral triangle.
BC = AC = AB (Length of all sides is same)
⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)
Also,
∠A + ∠B + ∠C = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
Therefore, ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

Page No: 128

Exercise 7.3

1. Î”ABC and Î”DBC 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) Î”ABD ≅ Î”ACD
(ii) Î”ABP ≅ Î”ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.

Given,
Î”ABC and Î”DBC are two isosceles triangles.

(i) In Î”ABD and Î”ACD,
AB = AC (Î”ABC is isosceles)
BD = CD (Î”DBC is isosceles)
Therefore, Î”ABD ≅ Î”ACD by SSS congruence condition.

(ii) In Î”ABP and Î”ACP,
AP = AP (Common)
∠PAB = ∠PAC (Î”ABD ≅ Î”ACD so by CPCT)
AB = AC (Î”ABC is isosceles)
Therefore, Î”ABP ≅ Î”ACP by SAS congruence condition.

(iii) ∠PAB = ∠PAC by CPCT as Î”ABD ≅ Î”ACD.
AP bisects ∠A. --- (i)
also,
In Î”BPD and Î”CPD,
PD = PD (Common)
BD = CD (Î”DBC is isosceles.)
BP = CP (Î”ABP ≅ Î”ACP so by CPCT.)
Therefore, Î”BPD ≅ Î”CPD by SSS congruence condition.
Thus, ∠BDP = ∠CDP by CPCT. --- (ii)
By (i) and (ii) we can say that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD (by CPCT as Î”BPD ≅ Î”CPD)
and BP = CP --- (i)
also,
∠BPD + ∠CPD = 180° (BC is a straight line.)
⇒ 2∠BPD = 180°
⇒ ∠BPD = 90° ---(ii)
From (i) and (ii),
AP is the perpendicular bisector of BC.

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

Given,
AD is an altitude and AB = AC

(i) In Î”ABD and Î”ACD,
AB = AC (Given)
Therefore, Î”ABD ≅ Î”ACD by RHS congruence condition.
Now,
BD = CD (by CPCT)

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 Î”PQR (see Fig. 7.40). Show that:
(i) Î”ABM ≅ Î”PQN
(ii) Î”ABC ≅ Î”PQR

Given,
AB = PQ, BC = QR and AM = PN

(i) 1/2 BC = BM and 1/2QR = QN (AM and PN are medians)
also,
BC = QR
⇒ 1/2 BC = 1/2QR
⇒ BM = QN
In Î”ABM and Î”PQN,
AM = PN (Given)
AB = PQ (Given)
BM = QN (Proved above)
Therefore, Î”ABM ≅ Î”PQN by SSS congruence condition.

(ii) In Î”ABC and Î”PQR,
AB = PQ (Given)
∠ABC = ∠PQR (by CPCT)
BC = QR (Given)

Therefore, Î”ABC ≅ Î”PQR by SAS congruence condition.

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

Given,
BE and CF are two equal altitudes.
In Î”BEC and Î”CFB,
∠BEC = ∠CFB = 90° (Altitudes)
BC = CB (Common)
BE = CF (Common)
Therefore, Î”BEC ≅ Î”CFB by RHS congruence condition.
Now,
∠C = ∠B (by CPCT)
Thus, AB = AC as sides opposite to the equal angles are equal.

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

Given,
AB = AC
In Î”ABP and Î”ACP,
∠APB = ∠APC = 90° (AP is altitude)
AB = AC (Given)
AP = AP (Common)
Therefore, Î”ABP ≅ Î”ACP by RHS congruence condition.
Thus, ∠B = ∠C (by CPCT)

Page No: 132

Exercise 7.4

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

ABC is a triangle right angled at B.
Now,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠C = 90° and ∠B is 90°.
Since, B is the largest angle of the triangle, the side opposite to it must be the largest.
So, BC is the hypotenuse which is the largest side of the right angled triangle ABC.

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

Given,
∠PBC < ∠QCB
Now,
∠ABC + ∠PBC = 180°
⇒ ∠ABC = 180° - ∠PBC
also,
∠ACB + ∠QCB = 180°
⇒ ∠ACB = 180° - ∠QCB
Since,
∠PBC < ∠QCB therefore, ∠ABC > ∠ACB
Thus, AC > AB as sides opposite to the larger angle is larger.

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

Given,
∠B < ∠A and ∠C < ∠D
Now,
AO <  BO --- (i) (Side opposite to the smaller angle is smaller)
OD < OC ---(ii) (Side opposite to the smaller angle is smaller)
AO + OD < BO + OC

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.

In Î”ABD,
∴ ∠ADB < ∠ABD --- (i) (Angle opposite to longer side is larger.)
Now,
In Î”BCD,
BC < DC < BD
∴ ∠BDC < ∠CBD --- (ii)
Adding (i) and (ii) we get,
∠ADB + ∠BDC < ∠ABD + ∠CBD
⇒ ∠B > ∠D
Similarly,
In Î”ABC,
∠ACB < ∠BAC --- (iii) (Angle opposite to longer side is larger.)
Now,
∠DCA < ∠DAC --- (iv)
Adding (iii) and (iv) we get,
∠ACB + ∠DCA < ∠BAC + ∠DAC
⇒ ∠A > ∠C

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

Given,
PR > PQ and PS bisects ∠QPR
To prove,
∠PSR > ∠PSQ
Proof,
∠PQR > ∠PRQ --- (i) (PR > PQ as angle opposite to larger side is larger.)
∠QPS = ∠RPS --- (ii) (PS bisects ∠QPR)
∠PSR = ∠PQR + ∠QPS --- (iii) (exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS --- (iv) (exterior angle of a triangle equals to the sum of opposite interior angles)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]

Page No: 133

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