Gujarat Board GSEB Solutions Class 9 Maths Chapter 7 Triangles Ex 7.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 7 Triangles Ex 7.1

Question 1.

In quadrilateral ABCD, AC = AD, and AB bisect ∠A (see in figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

Solution:

Given: Quadrilateral ABCD in which AB is the bisector of ∠A and AC = AD.

To show:

ΔABC ≅ ΔABD

Proof: In ΔABC and ΔABD,

AC = AD (Given)

∠BAC = ∠BAD

(AB is the bisector of ∠A)

AB = AB (Common)

∴ ΔABC ≅ ΔABD (by SAS Congruency)

Hence BC = BD (by CPCT)

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Question 2.

ABCD is a quadrilateral in which AD = BC and

∠DAB = ∠CBA (see figure). Prove that

1. ΔABD ≅ ΔBAC

2. BD = AC

3. ∠ABD = ∠BAC

Solution:

Given: ABCD is a quadrilateral in which

AD = BC and ∠DAB = ∠CBA.

1. In ΔABD and ΔBAC,

AD = BC (given)

∠DAB = ∠CBA (given)

AB = AB (common)

∴ ΔABD ≅ ΔBAC (by SAS congruency)

2. ΔABD ≅ ΔBAC (proved above)

∴ BD = AC (byCPCT)

3. ΔABD ≅ ΔBAC (proved above)

∴ ∠ABD = ∠BAC (by CPCT)

Question 3.

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

Solution:

Given: AD and BC are perpendiculars on AB

such that AD = BC.

To Prove: CD bisects AB.

i.e., OA = OD

Proof: In ΔAOD and ΔBOC,

AD = BC (Given)

∠OAD = ∠OBC (each 900)

AD || BC (If alternate interior angles are equal then lines are parallel)

∴ ∠ADO = ∠OCB (Alternate interior angle)

Hence ΔAOD ≅ ΔBOC (by ASA congruency)

Therefore, OA = OB (by CPCT)

∴ CD bisects AB.

Question 4.

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

ΔABC ≅ ΔCDA

Solution:

Given: l || m and both are intersected by another pair of parallel lines p and q

To Prove: ΔABC ≅ ΔCDA

Proof: Since AB || CD

and AD||BC

Now in ΔABC and ΔADC,

∠1 = ∠2 (Alternate interior angles

∠3 = ∠4 (Alternate interior angles)

and AC = AC (common)

ΔABC ≅ ΔADC (by ASA congruency)

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) ΔAPB ≅ ΔAQB

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

Solution:

Given: l is the bisector of ∠A, i.e., ∠BAP = ∠BAQ.

BP and BQ are perpendiculars from B to arms of ∠A.

(i) In ΔAPB and ΔAQB

∠BPA = ∠BQA (each 90°)

∠BAP = ∠BAQ (l is the angle bisector of ∠A)

and AB = AB (Common)

Hence ΔAPB ≅ ΔAQB (By AAS congruency)

(ii) Since ΔAPB ≅ ΔAQB

∴ BP = BP (By CPCT)

Question 6.

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

Solution:

Given: In figure AC = AE.

and AB = AD

and ∠BAD = ∠EAC

To show: BC = DE

Proof: In ΔABC and ΔADE,

AB = AD (given)

∠BAD = ∠EAC (given)

Adding ∠DAC on both sides, we get

∠BAD + ∠DAC = ∠EAC + ∠DAC

⇒ ∠BAC = ∠DAE

AC = AE (given)

∴ ΔBAC ≅ ΔDAE (By SAS congruency)

Hence BC = DE (By CPCT)

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) ADAP AEBP

(ii) AD = BE

Solution:

Proof: (i) In ΔDAP and ΔEBP,

∠BAD = ∠ABE (given)

or ∠DAP = ∠PBE

AP = PB (P is the mid≅points)

∠EPA = ∠DPB (given)

Adding ∠DPE on both sides, we get

∠EPA + ∠DPE = ∠DPB + ∠DPE

∠DPA = ∠EPB

∴ ΔDAP ≅ ΔEBP (By ASA congruency)

(ii) ΔDAP ≅ ΔEBP

∴ AD = BE (by CPCT)

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:

1. ∠AMC ≅ ΔBMD

2. ∠DBC is a right angle.

3. ΔDBC ≅ ΔACB

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

Solution:

Proof:

1. In ΔAMC and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMC = ∠BMD (Vertically opposite angle)

CM = DM (given)

Hence, ΔAMC ≅ ΔBMD (by SAS congruency)

AC = BD (by CPCT)

2. ΔAMC ≅ ΔBMD (Proved in part i)

∴ ∠ACM = ∠BDM (by CPCT)

(Alternate interior angles)

Hence AC || BD (If alternate interior angles are equal then lines are parallel)

∴ ∠ACB + ∠DBC = 180°

(Sum of consecutive interior angles is equal to 180°

∴ ∠DBC = 180°≅ 90°

⇒ ∠DBC = 90°

Hence ∠DBC is a right angle.

3. Now in ΔACB and ΔDBC,

∠ACB = ∠DBC (each 900)

BC = CB (common)

AC = BD (Proved in part i)

Hence ΔACB ≅ ΔDBC (by SAS congruency)

4. ΔDBC ≅ ΔACB (Proved in part iii)

DC = AB (by CPCT)

⇒ 2CM = AB (DM = CM)

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