Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.1 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 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?

Solution:

Now, in As ABC and ABD, we have

AC = AD [Given]

∠CAB = ∠BAD (∵ AB bisects ∠A]

and, AB = AB [Common]

∴ By SAS congruence criterion, we have

∆ ABC ≅ ∆ ABD

⇒ BC = BD

[∵ 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.

Solution:

In ∆s ABD and BAC, we have

AD = BC

∠DAB = ∠CBA lGiven]

AB = AB [Common]

By SAS criterion of congruence, we have A ABD = A BAC, which proves (i)

⇒ BD = AC

and, ∠ABD = ∠BAC, which proves (ii) and (iii)

[∵ Corresponding parts of congruent triangles are equal]

Question 3.

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

Solution:

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)]

∠D AO = ∠OBG [Each = 90°]

and, AD = BC [Given]

∴ By AAS congruence criterion, we have ∆ AOD ≅ ∆ BOC

OA = OB

[∵ Corresponding parts of congruent triangles are equal] i.e., O is the mid-point AB.

Hence, CD bisects AB.

Question 4.

1 and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ ABC ≅ ∆ CDA.

Solution:

Since l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore, AD || BC and AB || CD ⇒ ABCD is a parallelogram.

i.e., AB = CD

and BC = AD

Now, in ∆s ABC and CDA, we have

AB = CD [Prove above]

BC = AD [Prove above]

and AC = AC [Common]

∴ By SSS 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.

Solution:

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

Question 6.

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

Solution:

AB = AD

∠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.

Solution:

We have, ∠EPA = ∠DPB

⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE’

⇒ ∠DPA = ∠EPB … (1)

Now, in ∆s EBP and DAP, we have

∠EPB = ∠DPA [From (1)]

BP = AP [Given]

and, ∠EBP = ∠DAP [Given]

So, by ASA criterion of congruence, we have

∆ EBP ≅ ∆ DAP

⇒ BE = AD i.e., AD = BE

[∵ 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.

Solution:

(i) In ∆s AMC and BMD, we have

AM = BM [∵ M is the mid-point of AB]

∠AMC = ∠BMD [Vertically opp. ∠s]

and, CM = MD [Given]

∴ By SAS criterion of congruence, we have

∴ ∆ AMC ≅ ∆ BMD

(ii) Now, ∆ AMC ≅ ∆ BMD

⇒ BD = CA and ∠BDM = ∠ACM … (1)

[∵Corresponding parts of congruent triangles are equal]

Thus, transversal CD cuts CA and BD at C and D respectively such’that the alternate angles ∠BDM and ∠ACM are equal. Therefore, BD || CA.

⇒ ∠CBD + ∠BCA = 180°

[∵ Sum of the interior angles on the same side of ” transversal = 180°]

⇒ ∠CBD + 90° = 180° [∵ ∠BCA = 90°]

⇒ ∠DBC = 90°.

(iii) Now, in ∆s DBC and ACB, we have

BD = CA [From (1)]

∠DBC = ∠ACB [∵ Each = 90°)

BC = BC [Common]

By SAS criterion of congruence, we have ∆ DBC ≅ ∆ ACB.

(iv) CD = AB

[∵ Corresponding parts of congruent triangles are equal]

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