# Class 9 RD Sharma Solutions – Chapter 10 Congruent Triangles- Exercise 10.1

### Question 1. In the figure, the sides BA and CA have been produced such that: BA = AD and CA = AE. Prove that segment DE || BC.

Solution:

Given that:

BA = AD and CA = AE

Prove that DE || BC

So, here we consider triangle BAC and DAE,

CA= AE          -(given)

And âˆ BAC = âˆ DAE           -(vertically opposite angles)

Hence, by the SAS congruence criterion, we have

Î” BAC â‰ƒ Î”DAE

So, we can say that:

âˆ DEA = âˆ BCA, âˆ EDA = âˆ CBA          -(Corresponding parts of congruent triangles are equal)

Now, lines DE and BC are intersected by a transversal DB such that âˆ DEA = âˆ BCA.

Hence, proved that DE || BC.

### Question 2. In a Î” QPR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR, and RP respectively. Prove that LN = MN.

Solution:

Given that,

In Î” QPR, PQ = QR

Also, L, M, N are midpoints of the sides PQ, QP, and RP.

And we are given to prove that LN = MN.

In Î” QPR, PQ = QR, and L, M, N are midpoints

Hence, Î” QPR is an isosceles triangle.

PQ = QR

âˆ  QPR = âˆ  QRP          -(1)

And also, L and M are midpoints of PQ and QR.

Hence, we can say that:

PQ = QR

PL = LQ = QM = MR = PQ/2 = QR/2

Now, we have Î” LPN and Î” MRN,

LP = MR

âˆ LPN = âˆ MRN           -(From above)

âˆ QPR and âˆ LPN are the same.

And also âˆ QRP and âˆ MRN are the same.

PN = NR          -(N is the midpoint of PR)

So, by SAS congruence criterion, we can say that Î”LPN = Î”MRN

By corresponding parts of congruent triangles are equal we can say that LN = MN.

### Question 3. In the figure given, PQRS is a square and SRT is an equilateral triangle. Prove that (i) PT = QT (ii) âˆ TQR = 15Â°

Solution:

We are given that PQRS is a square and SRT is an equilateral triangle.

And we have to prove that

(i) PT = QT

PQRS is a square and SRT is an equilateral triangle.

Now we see that PQRS is a square

PQ = QR = RS = SP        -(1)

And âˆ SPQ = âˆ PQR = âˆ QRS = âˆ RSP = 90Â° = right angle

And also, SRT is an equilateral triangle.

SR = RT = TS          -(2)

And âˆ TSR = âˆ SRT = âˆ RTS = 60Â°

From equation(1) and (2) we can say that,

PQ = QR = SP = SR = RT = TS         -(3)

And also for angles we have,

âˆ TSP = âˆ TSR + âˆ RSP = 60Â° + 90Â° + 150Â°

âˆ TRQ = âˆ TRS + âˆ SRQ = 60Â° + 90Â° + 150Â°

âˆ TSR = âˆ TRQ = 150Â°          -(4)

SP = RQ           -(From eq (3))

Hence, by SAS congruence criterion we can say that Î”TSP = Î”TRQ i.e both the triangles are congruent.

PT = QT          (Corresponding parts of congruent triangles are equal)

(ii) âˆ TQR = 15Â°

Let’s see Î”TQR.

QR = TR

And also it is given that Î”TQR is an isosceles triangle.

âˆ QTR = âˆ TQR         -(Angles opposite to equal sides)

As we know that the sum of angles in a triangle is equal to 180âˆ˜

âˆ QTR + âˆ TQR + âˆ TRQ = 180Â°

2âˆ TQR + 150Â° = 180Â°

2âˆ TQR = 180Â° – 150Â°

2âˆ TQR = 30Â°

Hence, âˆ TQR = 15Â°

Hence, proved

### Question 4. Prove that the medians of an equilateral triangle are equal.

Solution:

Prove that the medians of an equilateral triangle are equal.

So, let D, E, F are midpoints of BC, CA, and AB. AD, BE and CF are medians of ABC.

Then, AD, BE and CF are medians of ABC.

Now, we can say that

D Is the midpoint of BC

So now we can say that,

BD = DC = BC/2

CE = EA = AC/2

AF = FB = AB/2

Since, it is already given that Î”ABC is an equilateral triangle

Therefore

AB = BC = CA

Hence, we can imply that,

BD = DC = CE = EA = AF = FB = BC/2 = AC/2 = AB/2

Also, we can say that âˆ ABC = âˆ BCA = âˆ CAB = 60Â°        -(Angles of equilateral triangle)

Now, consider Î”ABD and Î”BCE AB = BC

Here we find that BD = CE

Now, in Î”TSR and Î”TRQ

TS = TR

âˆ ABD = âˆ BCE

So, from SAS congruence criterion, we can conclude that Î”ABD = Î”BCE

Now, consider Î”BCE and Î”CAF, BC = CA

âˆ BCE = âˆ CAF       (From above)

CE = AF

So, from SAS congruence criterion, we have, Î”BCE = Î”CAF

BE = CF [from above]

Hence proved

### Question 5. In Î”ABC, if âˆ A = 120Â° and AB = AC. Find âˆ B and âˆ C.

Solution:

In the triangle ABC

Given that âˆ A = 120Â° and AB = AC

According to the question the triangle is isosceles

hence the angles opposite to equal sides are equal

âˆ B = âˆ C

We also know that sum of angles in a triangle is equal to 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

2âˆ B = 180Â° – 120Â°

âˆ B = âˆ C = 30Â°

### Question 6. In Î”ABC, if AB = AC and âˆ B = 70Â°. Find âˆ A.

Solution:

In the triangle ABC

Given that âˆ B = 70Â° and AB = AC

According to the question the triangle is isosceles

hence angles opposite to equal sides are equal.

âˆ B = âˆ C

Hence, âˆ C = 70Â°

We also know that sum of angles in a triangle is equal to 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

âˆ A = 40Â°

Hence, âˆ A = 40Â° and âˆ C = 70Â°

### Question 7. The vertical angle of an isosceles triangle is 100Â°. Find its base angles.

Solution:

Consider an isosceles Î”ABC such that AB = AC

Given that vertical angle, âˆ A = 100Â°

To find the base angles

Since Î”ABC is an isosceles triangle hence âˆ B = âˆ C

We know that sum of interior angles of a triangle = 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

100Â° + âˆ B +âˆ B = 180Â°

2âˆ B = 180Â° – 100Â°

âˆ B = 40Â°

âˆ B = âˆ C = 40Â°

Hence, âˆ B = 40Â° and âˆ C = 40Â°

### Question 8. In âˆ†ABC AB = AC and âˆ ACD = 105Â°. Find âˆ BAC.

Solution:

Given: AB = AC and âˆ ACD = 105Â°

Since, âˆ BCD = 180Â° = Straight angle

âˆ BCA + âˆ  ACD = 180Â°

âˆ BCA + 105Â° = 180Â°

Hence, âˆ BCA = 75Â°

Also, Î”ABC is an isosceles triangle

AB = AC

âˆ ABC = âˆ  ACB = 75Â°.

Now the Sum of Interior angles of a triangle = 180Â°

âˆ ABC = âˆ BCA + âˆ CAB = 180Â°

75Â° + 75Â° + âˆ CAB = 180Â°

âˆ BAC = 30Â°

### Question 9. Find the measure of each exterior angle of an equilateral triangle.

Solution:

We know that for an equilateral triangle

âˆ ABC = âˆ BCA = CAB =180Â°/3 = 60Â°

Now,

Extend side BC to D, CA to E, and AB to F.

âˆ BCA + âˆ ACD = 180Â°

60Â° + âˆ ACD = 180Â°

âˆ ACD = 120Â°

Similarly, we can say that, âˆ BAE = âˆ FBC = 120Â°

So, the measure of each exterior angle of an equilateral triangle = 120Â°.

### Question 10. If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.

Solution:

ED is a straight line segment and B and C are the points on it.

âˆ EBC = âˆ BCD          -(Both are equal to 180Â°)

âˆ EBA + âˆ ABC = âˆ ACB + âˆ ACD

Since âˆ ABC = âˆ ACB

Hence, âˆ EBA = âˆ ACD

Hence, proved

### Question 11. In the given Figure AB = AC and DB = DC, find the ratio âˆ ABD:âˆ ACD.

Solution:

Given: AB = AC, DB = DC

âˆ ABD = âˆ ACD

Hence, Î”ABC and Î”DBC are isosceles triangles

âˆ ABC = âˆ ACB and also âˆ DBC = âˆ DCB    (Angles opposite to equal sides are equal)

Now we have to find the ratio =âˆ ABD : âˆ ACD

âˆ ABD = (âˆ ABC – âˆ DBC)

âˆ ACD = (âˆ ACB – âˆ DCB)

(âˆ ABC – âˆ DBC):(âˆ ACB – âˆ DCB)

(âˆ ABC – âˆ DBC):(âˆ ABC – âˆ DBC)

1:1

Hence, âˆ  ABD:âˆ  ACD = 1:1

### Question 12. Determine the measure of each of the equal angles of a right-angled isosceles triangle.

Solution:

ABC is a right-angled triangle

âˆ A = 90Â° and AB = AC

Since,

AB = AC

âˆ C = âˆ B

As we know that the sum of angles in a triangle = 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

90Â° + âˆ  B+ âˆ  B = 180Â°

2âˆ B = 90Â°

âˆ B = 45Â°

âˆ B = 45Â°, âˆ C = 45Â°

So, the measure of each of the equal angles of a right-angled Isosceles triangle = 45Â°

### Question 13. AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from points A and B. Show that the line PQ is the perpendicular bisector of AB.

Solution:

Given:

AB is a line segment and P, Q are points on opposite sides of AB such that

AP = BP

AQ = BQ

Prove that PQ is the perpendicular bisector of AB.

Let Î”PAQ and Î”PBQ,

AP = BP

AQ = BQ

PQ – PQ

Î” PAQ â‰ƒ Î” PBQ are congruent by SAS congruent condition.

Now, we can observe that APB and ABQ are isosceles triangles.

âˆ  PAB = âˆ  ABQ

And also âˆ  QAB = âˆ  QBA

Now consider Î” PAC and Î” PBC

C is the point of intersection of AB and PQ

PA = PB

âˆ APC = âˆ BPC

PC = PC

So, by SAS congruency of triangle Î”PAC â‰… Î”PBC.

AC = CB and âˆ PCA = âˆ PBC

And also, ACB is the line segment

âˆ ACP + âˆ  BCP = 180Â°

âˆ ACP = âˆ PCB

âˆ ACP = âˆ PCB = 90Â°

We have AC = CB

C is the midpoint of AB

So, we can conclude that PC is the perpendicular bisector of AB and C is a point on the line PQ.

Therefore, PQ is the perpendicular bisector of AB.

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