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# Class 9 NCERT Solutions- Chapter 7 Triangles – Exercise 7.1

• Last Updated : 28 Dec, 2020

### Question 1. In quadrilateral ACBDAC = AD and AB bisects âˆ  A (see Fig. 7.16). Show that âˆ† ABC â‰… âˆ† ABD What can you say about BC and BD?

Solution:

Given that AC and AD are equal

i.e. AC = AD and the line AB bisects âˆ A.

Considering the two triangles Î”ABC and Î”ABD,

Where,

AC = AD { As given}………………………………………… (i)

âˆ CAB = âˆ DAB ( As AB  bisects of âˆ A)……………. (ii)

AB { Common side of both the triangle} …….. …(iii)

From above three equation both the triangle satisfies “SAS” congruency criterion

So, Î”ABC â‰… Î”ABD.

Also,

BC and BD will be of equal lengths as they are corresponding parts of congruent triangles(CPCT).

So BC = BD.

### Question 2. ABCD is a quadrilateral in which AD = BC and âˆ  DAB = âˆ  CBA (see Fig. 7.17). Prove that

(i) âˆ† ABD â‰… âˆ† BAC

(ii) BD = AC

(iii) âˆ  ABD = âˆ  BAC.

Solution:

(i) Given that AD = BC,

And âˆ  DAB = âˆ  CBA.

Considering two triangles Î”ABD and Î”BAC.

Where,

AD = BC { As given }………………………………………….. (i)

âˆ  DAB = âˆ  CBA { As given also}……………………….. (ii)

AB  {Common side of both the triangle)…………. (iii)

From above three equation two triangles ABD and BAC satisfies “SAS” congruency criterion

So, Î”ABD â‰… Î”BAC

(ii) Also,

BD and AC  will be equal as they are corresponding parts of congruent triangles(CPCT).

So BD = AC

(iii) Similarly,

âˆ ABD and âˆ BAC will be equal as they are corresponding parts of congruent triangles(CPCT).

So,

âˆ ABD = âˆ BAC.

### Question 3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Solution:

Given that AD and BC are two equal perpendiculars to a line segment AB

Considering two triangles Î”AOD and Î”BOC

Where,

âˆ  AOD = âˆ  BOC {Vertically opposite angles}………………. (i)

âˆ  OAD = âˆ  OBC {Given that they are perpendiculars}…. (ii)

AD = BC {As given}………………………………………………… (iii)

From above three equation both the triangle satisfies “AAS” congruency criterion

So, Î”AOD â‰… Î”BOC

AO and BO will be equal as they are corresponding parts of congruent triangles(CPCT).

So, AO = BO

Hence, CD bisects AB at O.

### Question 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that âˆ† ABC â‰… âˆ† CDA.

Solution:

Given that l and m are two parallel lines p and q are another pair of parallel lines

Considering two triangles Î”ABC and Î”CDA

Where,

âˆ  BCA = âˆ DAC  {Alternate interior angles}…. (i)

âˆ  BAC = âˆ  DCA {Alternate interior angles}…. (ii)

AC  {Common side of two triangles}………….(iii)

From above three equation both the triangle satisfies “ASA” congruency criterion

So, Î”ABC â‰… Î”CDA

### 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 Fig. 7.20). Show that:

(i) âˆ† APB â‰… âˆ† AQB

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

Solution:

Given that, Line l is the bisector of an angle âˆ  A and B

BP and BQ are perpendiculars from angle A.

Considering two triangles Î”APB and Î”AQB

Where,

âˆ  APB = âˆ  AQB { Two right angles as given }…… (i)

âˆ BAP = âˆ BAQ (As line l  bisects  angle A }……… (ii)

AB  { Common sides of both the triangle }……… (iii)

From above three equation both the triangle satisfies “AAS” congruency criterion

So, Î”APBâ‰… Î”AQB.

(ii) Also we can say BP and BQ are equal as they are corresponding parts of congruent triangles(CPCT).

So, BP = BQ

### Question 6. In Fig. 7.21, AC = AE, AB = AD and âˆ  BAD = âˆ  EAC. Show that BC = DE.

Solution:

Given that AC = AE, AB = AD

And âˆ BAD = âˆ EAC

As given that âˆ BAD = âˆ EAC

Adding âˆ DAC on both the sides

We get,

âˆ BAD + âˆ DAC = âˆ EAC + âˆ  DAC

âˆ BAC = âˆ EAD

Considering two triangles Î”ABC and Î”ADE

Where,

AC = AE  { As given }…………………… (i)

âˆ BAC = âˆ EAD { Hence proven }…….. (ii)

AB = AD {As also given }……………….. (iii)

From above three equation both the triangle satisfies “SAS” congruency criterion

So, Î”ABC â‰… Î”ADE

(ii) Also we can say BC and  DE are equal as they are corresponding parts of congruent triangles(CPCT).

So that BC = DE.

### 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 Fig. 7.22). Show that:

(i) âˆ† DAP â‰… âˆ† EBP

(ii) AD = BE

Solution:

Given that P is the mid-point of line  AB, So AP = BP

Also, âˆ  BAD = âˆ  ABE and âˆ  EPA = âˆ  DPB

Now adding âˆ DPE on both the sides of two equal angle âˆ  EPA = âˆ  DPB

âˆ  EPA + âˆ  DPE = âˆ  DPB + âˆ DPE

Which implies that two angles âˆ  DPA = âˆ  EPB

Considering two triangles âˆ† DAP and âˆ† EBP

âˆ  DPA = âˆ  EPB { Hence proven }…… (i)

AP = BP { Hence Given }……………… (ii)

âˆ  BAD = âˆ  ABE { As given }…………..(iii)

From above three equation both the triangle satisfies “ASA” congruency criterion

So, Î”DAP â‰… Î”EBP

(ii) Also we can say AD and BE are equal as they are corresponding parts of congruent triangles(CPCT).

So that, AD = BE

### 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 Fig. 7.23). Show that:

(i) âˆ† AMC â‰… âˆ† BMD

(ii) âˆ  DBC is a right angle.

(iii) âˆ† DBC â‰… âˆ† ACB

(iv) CM =  1/2 AB

Solution:

Given that M is the mid-point of  AB

So AM = BM

âˆ  ACB = 90Â°

and DM = CM

(i) Considering two triangles Î”AMC and Î”BMD:

AM = BM { As given }………………………………………. (i)

âˆ  CMA = âˆ  DMB { Vertically opposite angles }…. (ii)

CM = DM { As given also}……………………………….. (iii)

From above three equation both the triangle satisfies “SAS” congruency criterion

So, Î”AMC â‰… Î”BMD

(ii) From above congruency we can say

âˆ  ACD = âˆ  BDC

Also alternate interior angles of two parallel lines AC and DB.

Since sum of two co-interiors angles results to 180Â°.

So, âˆ  ACB + âˆ  DBC = 180Â°

âˆ  DBC = 180Â° – âˆ  ACB

âˆ  DBC = 90Â° { As âˆ ACB =90Â° }

(iii) In Î”DBC and Î”ACB,

BC  { Common side of both the triangle }……. (i)

âˆ  ACB = âˆ  DBC { As both are right angles }….(ii)

DB = AC (by CPCT)………………………………….. (iii)

From above three equation both the triangle satisfies “SAS” congruency criterion

So, Î”DBC â‰… Î”ACB

(iv) As M is the mid point so we can say

DM = CM = AM = BM

Also we can say that AB = CD ( By CPCT )

As M is the mid point of CD we can write

CM + DM = AB

Hence, CM + CM = AB  (As DM = CM )

Hence, CM = (Â½) AB

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