Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Textbook Exercise Questions and Answers.

## RBSE Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.1

Question 1.

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

Answer:

Now, in ∆s ABC and ABD, we have :

AC = AD (Given)

∠CAB = ∠BAD (∵ AB bisects ∠∆)

and, AB = AB (Common)

∴ By SAS congruence criterion, we have:

∆ABC ≅ ∆ABD

⇒ BC = BD, i.e. they are equal.

(∵ Corresponding parts of congruent triangles are equal.)

Question 2.

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that:

(i) ∆ABD ≅ ∆BAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

Answer:

In ∆s ABD and BAC, we have :

AD = BC (Given)

∠DAB = ∠CBA (Given)

AB = BA (Common)

∴ By SAS criterion of congruence, we have

∆ABD ≅ ∆BAC, which proves (i)

From (i), (ii) BD = AC (C.P.C.T.)

and, (iii) ∠ABD = ∠BAC

(∵ Corresponding parts of congruent triangles (CPCT) are equal.)

Question 3.

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

Answer:

Since AB and CD intersect at O, therefore

∠AOD = ∠BOC ….(1)

(Vertically opp. angles)

In ∆s AOD and BOC, we have:

∠AOD = ∠BOC [From (1)]

∠DAO = ∠CBO (Each = 90°)

and, AD = BC (Given)

∴By AAS congruence criterion, we have :

∆AOD ≅ ∆BOC

⇒ OA = OB

(∵ Corresponding parts of congruent triangles (CPCT) are equal.)

i.e. O is the mid-point AB.

Hence, CD bisects AB.

Question 4.

l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure).

Show that ∆ABC ≅ ∆CDA.

Answer:

Since l and m are two parallel lines intersected by another pair of parallel lines p and q, therefore AD || BC and AB || CD.

AB || CD and AC is a transversal.

∠BAC = ∠DCA (Alternate angles) ……….. (1)

Again, BC || AD and AC is a transversal.

∠BCA = ∠DAC (Alternate angles) ……….. (2)

Now, in ∆s ABC and CDA, we have :

∠BAC = ∠DCA [Proved above in (1)]

∠BCA = ∠DAC [Proved above in (2)]

and AC = CA

∴ By ASA criterion of congruence,

∆ABC ≅ ∆CDA.

Question 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 figure). Show that:

(i) ∆APB ≅ ∆AQB

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

Answer:

In ∆s APB and AQB, we have:

∠APB = ∠AQB (∵ Each = 90°)

∠PAB = ∠QAB (∵ AB bisects ∠PAQ)

AB = AB (Common)

By AAS congruence criterion, we have:

∆APB ≅ ∆AQB, which proves (i)

⇒ BP = PQ

(∵ Corresponding parts of congruent triangles are equal.)

i.e. B is equidistant from the arms of ∠A, which proves (ii).

Question 6.

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

Answer:

In ∆s ABC and ADE, we have :

AB = AD (Given)

∠BAC = ∠DAE

∵ ∠BAD = ∠EAC => ∠BAD + ∠DAC =

∠EAC + ∠DAC => ∠BAC = ∠DAE

and, AC = AE (Given)

∴ By SAS criterion of congruence, we have :

∆ABC ≅ ∆ADE

⇒ BC = DE

(∵ Corresponding parts of congruent triangles are equal.)

Question 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 figure). Show that:

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE

Answer:

We have: ∠EPA = ∠DPB

∴ ∠EPA + ∠DPE = ∠DPB + ∠DPE

or ∠DPA = ∠EPB

Now, in ∆s DAP and EBP, we have: …………. (1)

∠DPA = ∠EPB [From (1)]

AP = BP (Given)

and, ∠DAP = ∠EBP (Given)

So, by ASA criterion of congruence, we have:

∆DAP ≅ ∆EBP, which proves (i)

So, AD = BE, which proves (ii)

(∵ Corresponding parts of congruent triangles are equal.)

Question 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 figure). Show that:

(i) ∆AMC ≅ ∆BMD

(ii) ∠DBC is a right angle.

(iii) ∆DBC ≅ ∆ACB

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

Answer:

Given : ∠C = 90°, AM = BM, DM = CM

To prove:

(i) ∆AMC = ∆BMD

(ii) ∆DBC = 90°

(iii) ∆DBC = AACB

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

Proof:

(i) In ∆AMC and ∆BMD

AM = BM

MC = MD

∠1 = ∠2

∴ ∆AMC ≅ ∆BMD (by SAS)

and so AC = DB and ∠4 = ∠3 (By CPCT).

(ii) ∠4 = ∠3 (Proved above)

∠4 = ∠3 But these are alternate interior angles.

∴ DB || AC

AS AC || BD and BC is the transversal.

∠C + ∠B = 180° (Cointerior angles)

⇒ ∠B = 180° – 90° = 90° (∵ ∠C = 90°)

(iii) In ∆DBC and ∆ACB

DB = AC (Proved)

∠B = ∠C (Each 90°)

BC = CB (Common)

∴ ∆DBC ≅ ∆ACB (By SAS)

and so DC =AB (By CPCT) (Proved)

(iv) DC = AB

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

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

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