RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

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.