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 âˆ A (see in figure). Show that Î”ABC â‰… Î”ABD. What can you say about BC and BD?

Solution:

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

To show:

Î”ABC â‰… Î”ABD

Proof: In Î”ABC and Î”ABD,

AC = AD (Given)

âˆ BAC = âˆ BAD

(AB is the bisector of âˆ A)

AB = AB (Common)

âˆ´ Î”ABC â‰… Î”ABD (by SAS Congruency)

Hence BC = BD (by CPCT)

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Question 2.

ABCD is a quadrilateral in which AD = BC and

âˆ DAB = âˆ CBA (see figure). Prove that

1. Î”ABD â‰… Î”BAC

2. BD = AC

3. âˆ ABD = âˆ BAC

Solution:

Given: ABCD is a quadrilateral in which

AD = BC and âˆ DAB = âˆ CBA.

1. In Î”ABD and Î”BAC,

AD = BC (given)

âˆ DAB = âˆ CBA (given)

AB = AB (common)

âˆ´ Î”ABD â‰… Î”BAC (by SAS congruency)

2. Î”ABD â‰… Î”BAC (proved above)

âˆ´ BD = AC (byCPCT)

3. Î”ABD â‰… Î”BAC (proved above)

âˆ´ âˆ ABD = âˆ 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 Î”AOD and Î”BOC,

AD = BC (Given)

âˆ OAD = âˆ OBC (each 900)

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

âˆ´ âˆ ADO = âˆ OCB (Alternate interior angle)

Hence Î”AOD â‰… Î”BOC (by ASA congruency)

Therefore, OA = OB (by CPCT)

âˆ´ 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

Î”ABC â‰… Î”CDA

Solution:

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

To Prove: Î”ABC â‰… Î”CDA

Proof: Since AB || CD

and AD||BC

Now in Î”ABC and Î”ADC,

âˆ 1 = âˆ 2 (Alternate interior angles

âˆ 3 = âˆ 4 (Alternate interior angles)

and AC = AC (common)

Î”ABC â‰… Î”ADC (by ASA congruency)

Question 5.

Line l is the bisector of an angle âˆ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of âˆ A (see figure). Show that

(i) Î”APB â‰… Î”AQB

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

Solution:

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

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

(i) In Î”APB and Î”AQB

âˆ BPA = âˆ BQA (each 90Â°)

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

and AB = AB (Common)

Hence Î”APB â‰… Î”AQB (By AAS congruency)

(ii) Since Î”APB â‰… Î”AQB

âˆ´ BP = BP (By CPCT)

Question 6.

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

Solution:

Given: In figure AC = AE.

and AB = AD

and âˆ BAD = âˆ EAC

To show: BC = DE

Proof: In Î”ABC and Î”ADE,

AB = AD (given)

âˆ BAD = âˆ EAC (given)

Adding âˆ DAC on both sides, we get

âˆ BAD + âˆ DAC = âˆ EAC + âˆ DAC

â‡’ âˆ BAC = âˆ DAE

AC = AE (given)

âˆ´ Î”BAC â‰… Î”DAE (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 âˆ BAD = âˆ ABE and âˆ EPA = âˆ DPB (see figure). Show that.

(i) ADAP AEBP

(ii) AD = BE

Solution:

Proof: (i) In Î”DAP and Î”EBP,

âˆ BAD = âˆ ABE (given)

or âˆ DAP = âˆ PBE

AP = PB (P is the midâ‰…points)

âˆ EPA = âˆ DPB (given)

Adding âˆ DPE on both sides, we get

âˆ EPA + âˆ DPE = âˆ DPB + âˆ DPE

âˆ DPA = âˆ EPB

âˆ´ Î”DAP â‰… Î”EBP (By ASA congruency)

(ii) Î”DAP â‰… Î”EBP

âˆ´ 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. âˆ AMC â‰… Î”BMD

2. âˆ DBC is a right angle.

3. Î”DBC â‰… Î”ACB

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

Solution:

Proof:

1. In Î”AMC and Î”BMD,

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

âˆ AMC = âˆ BMD (Vertically opposite angle)

CM = DM (given)

Hence, Î”AMC â‰… Î”BMD (by SAS congruency)

AC = BD (by CPCT)

2. Î”AMC â‰… Î”BMD (Proved in part i)

âˆ´ âˆ ACM = âˆ BDM (by CPCT)

(Alternate interior angles)

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

âˆ´ âˆ ACB + âˆ DBC = 180Â°

(Sum of consecutive interior angles is equal to 180Â°

âˆ´ âˆ DBC = 180Â°â‰… 90Â°

â‡’ âˆ DBC = 90Â°

Hence âˆ DBC is a right angle.

3. Now in Î”ACB and Î”DBC,

âˆ ACB = âˆ DBC (each 900)

BC = CB (common)

AC = BD (Proved in part i)

Hence Î”ACB â‰… Î”DBC (by SAS congruency)

4. Î”DBC â‰… Î”ACB (Proved in part iii)

DC = AB (by CPCT)

â‡’ 2CM = AB (DM = CM)

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