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## Karnataka State Syllabus Class 9 Maths Chapter 5 Triangles Ex 5.3

Question 1.

∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. 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:

Data: ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.

To Prove:

(i) ∆ABD ≅ ∆ACD

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC

(v) AD is the angular bisector of ∠A.

Proof:

(i) In ∆ABD and ∆ACD,

AB = AC (data)

BD = DC (data)

AD is common.

S.S.S. Congruence rule.

∴ ∆ABD ≅ ∆ACD

(ii) In ∆ABP and ∆ACP,

AB = AC (data)

∠ABP = ∠ACP (Opposite angles)

∠BAP = ∠CAP (∵ ∆ABD ≅ ∆ACD proved)

Now ASA postulate.

∆ABP ≅ ∆ACP.

(iii) ∆BAD ≅ ∆CAD proved.

AP bisects ∠A.

In ∆BDP and ∆CDP,

BD = DC (data)

BP = PC (proved)

DP is common.

∴ ∆BDP ≅ ∆CDP (SSS postulate)

∴ ∠BDP = ∠CDP

∴ DP bisects ∠D.

∴ AP bisects ∠D.

(iv) Now, ∠APB + ∠APC = 180° (Linear pair)

∠APB + ∠APB = 180°

2 ∠APB = 180

∴ ∠APB = \(\frac{180}{2}\)

∴∠APB = 90°

∠APB = ∠APC = 90°

BP = PC (proved)

∴ AP is the perpendicular bisector BC.

(v) AP is the angular bisector of ∠A.

Angular bisector of ∠A is aD, because AD, AP are in one line.

Question 2.

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

(i) AD bisects BC

(ii) AD bisects ∠A.

Solution:

Data: AD is an altitude of an isosceles triangle ABC in which AB = AC.

To Prove:

(i) AD bisects BC.

(ii) AD bisects ∠A.

Proof: i) In ∆ABD and ∆ACD,

∠ADB = ∠ADC (∵ AD ⊥ BC)

AB = AC (data)

AD is common.

∴ ∆ABD ≅ ∆ACD

∴ BD = DC

∴ AD bisects BC.

(ii) ∠BAD = ∠CAD (∵ ∆ADB ≅ ∆ADC)

∴ AD bisects ∠A.

Question 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 ∆PQR. Show that :

(i) ∆ABM ≅ ∆PQN

(ii) ∆ABC ≅ ∆PQR.

Solution:

Data: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR.

To Prove:

(i) ∆ABM ≅ ∆PQN

(ii) ∆ABC ≅ ∆PQR.

Proof: (i) In ∆ABC,

AM is the median drawn to BC.

∴ BM = \(\frac{1}{2} \) BC

Similarly, in ∆PQR,

QN = \(\frac{1}{2}\) QR

But, BC = QR

\(\frac{1}{2} \) BC = \(\frac{1}{2}\) QR

∴ BM = QN

In ∆ABM and ∆PQN,

AB = PQ (data)

BM = QN (data)

AM = PN (proved)

∴ ∆ABM ≅ ∆PQN (SSS postulate)

(ii) In ∆ABC and ∆PQR,

AB = PQ (data)

∠ABC = ∠PQR (proved)

BC = QR (data)

∴ ∆ABC ≅ ∆PQR (SSS postulate)

Question 4.

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

Solution:

Data: BE and CF are two equal altitudes of a triangle ABC.

To Prove: ABC is an isosceles triangle.

Proof : BE = CF (data)

In ∆BCF and ∆CBE,

∠BFC = ∠CEB = 90° (data)

BC is a common hypotenuse.

As per Right angle, hypotenuse, side postulate,

∴ ∆BCF ≅ ∆CBE

∴ ∠CBF = ∠BCE

∴ ∠CBA = ∠BCA

∴ AB = AC

∴ ∆ABC is an isosceles triangle.

Question 5.

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

Solution:

Data: ABC is an isosceles triangle with AB = AC.

To Prove : ∠B = ∠C

Construction: Draw AP ⊥ BC.

Proof: In ∆ABC, AP ⊥ BC and AB = BC.

∴ In ∆ABP and ∆ACP

∠APB = ∠APC = 90° ( ∵ AP ⊥ BC)

Hypotenuse AB = Hypotenuse AC

AP is common.

As per RHS Postulate,

∆ABP ≅ ∆ACP

∴ ∠ABP = ∠ACP

∴ ∠ABC = ∠ACB

∴∠B = ∠C.