**KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1** are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.1.

## Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.1

Question 1.

In quadrilateral ACBD, AC = AD, and AB bisect ∠A. Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

Solution:

Data: In a quadrilateral ABCD,

AC = AD and AB bisect ∠A.

To Prove: ∆ABC ≅ ∆ABD

Proof: In ∆ABC and ∆ABD

AC = AD (Data)

∠CAB = ∠DAB .

∠A bisected

AB is common.

Here Side, angle, side rule is there.

∴ ∆ABC ≅ ∆ABD.

Question 2.

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that

(1) ∆ABD ≅ ∆BAC

(2) BD = AC

(3) ∠ABD = ∠BAC A

Solution:

Data: ABCD is a quadrilateral.

AD = BC and ∠DAB = ∠CBA.

To Prove:

(1) ∆ABD ≅ ∆BAC

(2) BD = AC

(3) ∠ABD = ∠BAC

Proof: (i) In ∆ABD and ∆BAC

AD = BC (Data)

∠DAB = ∠CBA (Data)

AB is common SAS postulate.

∴ ∆ABD ≅ ∆BAC.

(ii) ∆ABD ≅ ∆BAC.

∵ Corresponding sides are equal.

∴ BD = AC.

(iii) ∆ABD ≅ ∆BAC. (Proved)

Equal sides of Adjacent angles are equal.

As we have AD = BC,

The adjacent angle for AD is ABD

The adjacent angle for BC is BAC

∴∠ABD = ∠BAC.

Question 3.

AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Solution:

Data: AD and BC are equal perpendiculars to a line segment AB.

To Prove: CD Bisects AB.

Proof: In ∆CBO and ∆DAO,

BC = AD

∠CBO = ∠DAO = 90°

∠BOC = ∠AOD (Vertically opposite angles)

SAS postulate,

∴ ∆CBO = ∆DAO ∴ OA = OB

∴ CD bisects AB at ‘O’.

Question 4.

l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

Solution:

Data: l and m are two parallel lines intersected by another pair of parallel lines p and q.

To Prove: ∆ABC ≅ ∆CDA

Proof: In AABC and ACDA,

∠ACB = ∠DAC [∵ Alternate angles.]

∠BAC = ∠ACD

AC is common.

A.S.A. postulate.

∆ABC ≅ ∆CDA

Question 5.

Line 1 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. Show that :

(i) ∆APB ≅ ∆AQB

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

Solution:

Data: 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.

To Prove:

(i) ∆APB ≅ ∆AQB

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

Proof: In ∆APB and ∆AQB

∠APB = ∠AQB = 90°

∠PAB = ∠QAB (∵ l bisects ∠A).

AB is common.

∴ ∆APB ≅ ∆AQB (ASA postulate)

∴ BP = BQ.

B is equidistant from the arms of ∠A.

Question 6.

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

Solution:

Data: AC = AE, AB = AD and ∠BAD = ∠EAC

To Prove : BC = DE

Proof: In ∆ABC and ∆EAD,

AB = AD (Data)

AC = AE (Data)

∆BAC = ∆EAD

(∵∠BAD + ∠DAC = AEAC + ∠DAC and ∠DAC = ∠DAC.)

Side – Angle – Side (SAS) Postulate

∴ ∆ABC = ∆EAD

∴ BC = DE.

Question 7.

AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE.

Solution:

Data: 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.

To Prove :

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE

Proof : (i) In ∆DAP and ∆EBP,

AP = BP (∵ P is the mid-point of AB)

∠BAD = ∠ABE (Data)

∠APD = ∠BPE

∵ (∠EPA = ∠DPB Adding ∠EPD to both sides)

∠EPA + ∠EPD = ∠DPB + ∠EPD

∴ ∠APD = -BPE.

Now, Angle, Side, Angle postulate.

∴ ∆DAP ≅ ∆EBP.

(ii) As it is ∆DAP ≅ ∆EBP,

Three sides and three angles are equal to each other.

∴ AD = BE.

Question 8.

In the 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. Show that :

(i) ∆AMC ≅ ∆BMD

(ii) ∠DBC is a right angle.

(iii) ∆DBC ≅ ∆ACB

(iv) CM = \(\frac{1}{2}\) AB.

Solution:

Data: In a 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.

To Prove:

(i) ∆AMC ≅ ∆BMD

(ii) ∠DBC is a right angle.

(iii) ∆DBC ≅ ∆ACB

(iv) CM = \(\frac{1}{2}\) AB.

Proof: (i) In ∆AMC and ∆BMD,

BM = AM (∵ M is the mid-point of AB)

DM = CM (Data)

∠BMD = ∠AMC (Vertically opposite angles)

Now, Side, Angle, Side postulate.

∴ ∆AMC ≅ ∆BMD

(ii) ∆AMC ≅ ∆BMD (Proved)

Adding AMBC to both sides,

∆BMD + ∆MBC = ∆AMC + ∆MBC

∆DBC ≅ ∆ACB.

∴ AB Hypotenuse = DC Hypotenuse

∠ACB = ∠DBC = 90°

∴ ∠DBC is a right angle.

(iii) In ∆DBC and ∆ACB,

DB = AC

∵ ∆DMB ≅ DMC (Proved)

∠DBC = ∠ACB (Proved)

BC is Common.

Side, angle, side postulate.

∴ ∠DBC ≅ ∆ACB.

(iv) ∆DBC ≅ ∆ACB (proved)

Hypotenuse AB = Hypotenuse DC

\(\frac{1}{2}\) AB = \(\frac{1}{2}\) DC

\(\frac{1}{2}\) AB = CM

∴ CM = \(\frac{1}{2}\)AB.

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.1, drop a comment below and we will get back to you at the earliest.