NCERT Solutions for Class 9 Math Chapter 7.3 Triangles

---

EXERCISE 7.1 EXERCISE 7.2 EXERCISE 7.4 EXERCISE 7.5*

---

1.    `\▵\ `ABC and `\▵\ `DBC are two isosceles triangle 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) `\▵ABD≅▵ACD\ `
(ii) `\▵ABP≅▵ACP\ `
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.

solution :

   Given,

   `\▵\ `ABC and `\▵\ `DBC are isosceles triangle.

   Find,

   (i) `\▵ABD≅▵ACD\ `

   (ii) `\▵ABP≅▵ACP\ `

   (iii) AP bisects ∠A as well as ∠D.

   (iv) AP is the perpendicular bisector of BC.

   In `\▵\ `ABD and `\▵\ `ACD,

   AB = AC[ Given ]

   BD = CD[ Given ]

   AD = AD[ Common ]

   (i) `\▵ABD ≅ ▵ ACD\ `[ Rule SSS ]

   ∠BAD = ∠CAD `\-----(I)\ `[ CPCT ]

   Now,

   In `\▵\ `ABP and `\▵\ `ACP,

   AB = AC[ Given ]

   ∠BAP = ∠CAP[ From (I) ]

   AP = AP[ Common ]

   (ii) `\▵ABP ≅ ▵ ACP\ `[ Rule SAS ]

   BP = CP `\-----(II)\ `[ CPCT ]

   Similarly,

   In `\▵\ `BDP and `\▵\ `CDP,

   BD = CD[ Given ]

   DP = DP[ Common ]

   BP = CP[ From (II) ]

   `\▵BDP ≅ ▵ CDP\ `[ Rule SSS ]

   ∠BDP = ∠CDP `\-----(III)\ `[ CPCT ]

   From (I) and (III),

   we say,

   (iii) AP bisects ∠A as well as ∠D.

   ∠BPD + ∠CPD = 180°[ Linear pair of angles ]

   ⇒ ∠BPD + ∠BPD = 180°[ From (III) ]

   ⇒ 2 ∠BPD = 180°

   ⇒ ∠BPD = `\frac{180°}{2}\ `

   ⇒ ∠BPD = 90°

   As well as ∠CPD = 90°

   From (II),

   we say,

   (iv) 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 :

   Given,

   AD ⊥ BC and AB = BC

   Find,

   (i) AD bisects BC

   (ii) AD bisects ∠A

   In `\▵\ `ADB and `\▵\ `ADC

   AB = AC[ Given ]

   AD = AD[ Common ]

   ∠ADB = ∠ADC[ Given 90° ]

   `\▵ADB≅▵ADC\ `[ Rule RHS ]

   AD = CD[ CPCT ]

   we say,

   (i) AD bisects BC

   ∠BAD = ∠CAD[ CPCT ]

   we say,

   (ii) AD bisects ∠A.

3.    Two sides AB and BC median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of `\▵\ `PQR (see Fig. 7.40). Show that:
(i) `\▵ABM≅▵PQN\ `
(ii) `\▵ABC≅▵PQR\ `

solution :

   Given,

   AB = PQ, BC = QR, AM = PN, BC median AM and QR median PN.

   Find,

   (i) `\▵ABM≅▵PQN\ `

   (ii) `\▵ABC≅▵PQR\ `

   In Fig.,

   BC = QR

   `\=>frac{1}{2}BC=BM\ `[ BC median AM ]

   `\=>frac{1}{2}QR=QN\ `[ QR median PN ]

   we say,

   BM = QN `\-----(I)\ `

   In `\▵\ `ABM and `\▵\ `PQN

   AB = PQ[ Given ]

   AM = PN[ Given ]

   BM = QN[ From (I) ]

   (i) `\▵ABM≅▵PQN\ `[ Rule SSS ]

   ∠ABM = ∠PQN `\-----(II)\ `[ CPCT ]

   In `\▵\ `ABC and `\▵\ `PQR

   AB = PQ[ Given ]

   ∠ABC = ∠PQR[ From (II) ]

   BC = QR[ Given ]

   (ii) `\▵ABC≅▵PQR\ `[ Rule SAS ]

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

solution :

   Given,

   BE ⊥ AC, FC ⊥ AB and BE = FC

   Find,

   AB = AC

   In `\▵\ `BEC and `\▵\ `CFB

   ∠BEC = ∠CFB[ Given 90° ]

   BE = CF [ Given ]

   BC = BC [ Common ]

   `\▵BEC≅▵CFB\ `[ Rule RHS ]

   ∠B = ∠C[ CPCT ]

   AB = AC[ sides opposite to the equal angles ]

   So, we say that

   the triangle ABC is isosceles.

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

solution :

   Given,

   AB = AC and AP ⊥ BC

   Find,

   ∠B = ∠C

   In `\▵\ `APB and `\▵\ `APC

   AP = AP[ Common ]

   ∠APB = ∠APC[ Given 90° ]

   AB = AC[ Given ]

   `\▵APB≅▵APC\ `[ Rule RHS ]

   ∠B = ∠C[ CPCT ]

---

EXERCISE 7.1 EXERCISE 7.2 EXERCISE 7.4 EXERCISE 7.5*