RS Aggarwal Class 9 Solutions Chapter 5 - Congruence And Inequalities In Triangle Ex 5B (5.2)

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Question 1:

In the given figure, if x = y and AB = CB, then prove that AE = CD.

Solution:

Given: AB = BC

and, x^o = y^o

To prove: AE = CD

Proof: In \(\Delta\)ABE, we have,

Exterior \(\angle\)AEB = \(\angle\)EBA + \(\angle\)BAE

\(\Rightarrow\) y^o = \(\angle\)EBA + \(\angle\)BAE

Again, in \(\Delta\)BCD, we have

x^o = \(\angle\)CBA + \(\angle\)BCD

Since, x^o = y^o [Given]

So, \(\angle\)EBA + \(\angle\)BAE = \(\angle\)CBA + \(\angle\)BCD

\(\Rightarrow \) \(\angle\)BAE = \(\angle\)BCD

Thus in \(\Delta\)BCD and \(\Delta\)BAE, we have

\(\angle\)B = \(\angle\)B [Common]

BC = AB [Given]

and, \) \(\angle\)BAE = \(\angle\)BCD [Proved above]

Thus by Angle-Side-Angle criterion of congruence, we have

\(\Delta\)BCD \(\cong\) \(\Delta\)BAE

The corresponding parts of the congruent triangles are equal.

So, CD = AE [Proved]

Question 2:

ABC is a triangle in which AB = AC. If the bisectors of \(\angle\)B and \(\angle\)C meet AC and AB in D and E respectively, prove that BD = CE.

Solution:

Given: A \(\Delta\)ABC in which AB = AC and BD and CE are the bisectors of \(\angle\)B and \(\angle\)C respectively.

To prove: BD = CE

Proof: In \(\Delta\)ABC and \(\Delta\)ACE

\(\angle\)ABD = ½ \(\angle\)B

and, \(\angle\)ACE = ½ \(\angle\)C

But \(\angle\)B = \(\angle\)C as AB = AC [In isosceles triangle, base angles are equal]

\(\Rightarrow \) \(\angle\)ABD = \(\angle\)ACE

AB = AC [Given]

\(\angle\)A = \(\angle\)A [Common]

Thus by Angle-Side-Angle criterion of congruence, we have

\(\Delta\)ABD \(\cong\) \(\Delta\)ACE

The corresponding parts of the congruent triangles are equal.

BD = CE [c.p.c.t]

Question 3:

In the adjoining figure, AD is a median of \(\Delta\)ABC. If BL and CM are drawn perpendiculars on AD and AD produced, prove that BL = CM.

Solution:

Given: A triangle in which D is the midpoint of BC and BL \(\perp\) AD and CM \(\perp\) AD

To prove: BL = CM

Proof: In \(\Delta\)BLD and \(\Delta\)CMD

\(\angle\)BLD = \(\angle\)CMD = 90^o [Given]

\(\angle\)BDL = \(\angle\)MDC [Vertically opposite angles]

BD = DC [Given]

Thus by Angle-Angle-Side criterion of congruence, we have

\(\Delta\)BLD \(\cong\) \(\Delta\)CMD

The corresponding parts of the congruent triangles are equal

So, BL = CM [c.p.c.t]

Question 4:

In \(\Delta\)ABC, D is the mid-point of BC. If DL \(\perp\) AB and DM \(\perp\) AC such that DL = DM, prove that AB = AC.

Solution:

Given: In a \(\Delta\)ABC, D is the midpoint of BC and DL \(\perp\) AB and DM \(\perp\) AC. Also, DL = DM

To prove: AB = AC

Proof: In right angled triangle \(\Delta\)BLD and \(\Delta\)CMD

\(\angle\)BLD = \(\angle\)CMD = 90^o

Hypt. BD = Hypt. CD [Given]

DL = DM [Given]

Thus, by Right Angle-Hypotenuse_side criterion of congruence, we have

\(\Delta\)BLD \(\cong\) \(\Delta\)CMD

The corresponding parts of the congruent triangles are equal.

Therefore, \(\angle\)ABD = \(\angle\)ACD [c.p.c.t]

In \(\Delta\)ABC, we have

\(\angle\)ABD = \(\angle\)ACD

\(\Rightarrow \) AB = AC [Since opposite to equal angles are equal]

Question 5:

In \(\Delta\)ABC, AB = AC and the bisectors of \(\angle\)B and \(\angle\)C meet at a point O. Prove that BO = CO and the ray AO is the bisector of \(\angle\)A.

Solution:

Given: A \(\Delta\)ABC in which AB = AC, BO and CO are bisectors of \(\angle\)B and \(\angle\)C

To prove: In \(\Delta\)BOC, we have,

\(\angle\)OBC = ½ \(\angle\)B

and, \(\angle\)OCB = ½ \(\angle\)C

But, \(\angle\)B = \(\angle\)C [therefore AB = AC (given)]

So, \(\angle\)OBC = \(\angle\)OCB

Since base angles are equal, sides are equal

\(\Rightarrow \) OB = OC ——- (1)

Since OB and OC are the bisectors of angles,

\(\angle\)B and \(\angle\)C respectively, we have

\(\angle\)ABO = ½ \(\angle\)B

\(\angle\)ACO = ½ \(\angle\)C

\(\Rightarrow \) \(\angle\)ABO = \(\angle\)ACO  ——— (2)

Now, in \(\Delta\)ABO and \(\Delta\)ACO

AB = AC [Given]

\(\angle\)ABO = \(\angle\)ACO [from (2)]

BO = OC [from(1)]

Thus, by Side-Angle-Side criterion of congruence, we have

\(\Delta\)ABO \(\cong\) \(\Delta\)ACO

The corresponding parts of the congruent triangles are equal

Therefore, \(\angle\)BAO = \(\angle\)CAO [by c.p.c.t]

i.e. , AO bisects \(\angle\)A.

Question 6:

In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that

(i) PT = PS

(ii) \(\angle\)PSR = 15^o

Solution:

Given: PQR is an equilateral triangle and QRST is a square.

To prove: PT = PS

and \(\angle\)PSR = 15^o

Proof: Since \(\Delta\) is an equilateral triangle,

\(\angle\) = 60^o and \(\angle\)PRQ = 60^o

Since QRTS is a square,

\(\angle\)RQT = 90^o and \(\angle\)QRS = 90^o

In \(\Delta\)PQT

\(\angle\)PQT = \(\angle\)PQR + \(\angle\)RQT

= 60^o + 90^o = 150^o

In \(\Delta\)PRS

\(\angle\)PRS = \(\angle\)PRQ + \(\angle\)QRS

=60^o + 90^o = 150^o ——- (1)

\(\Rightarrow \) \(\angle\)PQT = \(\angle\)PRS ——— (2)

Thus, in \(\Delta\)PQT and \(\Delta\)PRS

PQ = RS [sides of equilateral triangle PQR]

\(\angle\)PQT = \(\angle\)PRS ——– (2)

QT = RS [sides of square QRST]

Thus, by Side-Angle-Side criterion of congruence, we have

therefore,  \(\Delta\)PQT \(\cong\) \(\Delta\)PRS

The corresponding parts of the congruent triangles are equal.

Therefore, PT = PS [c.p.c.t]

Now in \(\Delta\)PRS, we have

PR = RS

\(\Rightarrow \) \(\angle\) RPS = \(\angle\)PSR

But \(\angle\)PRS = 150^o [from(1)]

So, by angle sum property in \(\Delta\)PRS

\(\angle\)PRS + \(\angle\)SPR + \(\angle\)PSR = 180^o

\(\Rightarrow \) 150^o + \(\angle\)PSR + \(\angle\)PSR = 180^o

\(\Rightarrow \) 2\(\angle\)PSR = 180^o – 150^o

\(\Rightarrow \) \(\angle\)PSR = 30^o/2 = 15^o

Question 7:

In the given figure, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD = BF.

Solution:

Given: ABC is a triangle right angled at B. ACFG is a square and BCDE is a square.

To prove: AD = EF

Proof: Since BCDE is a square,

\(\angle\)BCD = 90^o ——- (1)

In \(\Delta\)ACD,

\(\angle\)ACD = \(\angle\)ACB + \(\angle\)BCD

= \(\angle\)ACB + 90^o ——- (2)

In \(\Delta\)BCF,

\(\angle\)BCF = \(\angle\)BCA + \(\angle\)ACF

Since ACFG is a square,

\(\angle\)ACF = 90^o

Thus, we have

\(\angle\)BCF = \(\angle\)BCA + 90^o ——- (3)

From (2) and (3), we have

\(\angle\)ACD = \(\angle\)BCF ——— (4)

Thus in \(\Delta\)ACD and \(\Delta\)BCF, we have

AC = CF [sides of a square]

\(\angle\)ACD = \(\angle\)BCF [from (4)]

CD = BC [sides of a square]

Thus, by Side-Angle-Side criterion of congruence, we have

Therefore, \(\Delta\)ACD \(\cong\) \(\Delta\)BCF

The corresponding parts of congruent triangles are equal.

So, AD = BF (c.p.c.t)

Question 8:

Prove that the median from the vertex of an isosceles triangle is the bisector of the vertical angle.

Solution:

Given: ABC is a isosceles triangle in which AB = AC and AD is the median through A.

To prove: \(\angle\)BAD = \(\angle\)DAC

Proof: In \(\Delta\)ABD and \(\Delta\)ADC

AB = AC [Given]

BD = DC [Given]

AD = AD [Common]

Thus by Side-Side-Side criterion of congruence, we have

\(\Delta\)ABD \(\cong\) \(\Delta\)ADC

The corresponding parts of the congruent triangles are equal.

Therefore, \(\angle\)BAD = \(\angle\)DAC (Proved)

Question 9:

In the given figure, ABCD is quadrilateral in which AB || DC and P is the mid-point of BC. On producing, AO and DC meet at Q. Prove that

(i) AB = CQ

(ii) DQ = DC + AB

Solution:

Given ABCD is a quadrilateral in which AB || DC

To prove:

i) AB = CQ

ii) DQ = DC + AB

Proof: In \(\Delta\)ABP and \(\Delta\)PCQ we have

\(\angle\)PAB = \(\angle\)PQC [alternate angles]

\(\angle\)APB = \(\angle\)CPQ [vertically opposite angles]

BP = PC [Given]

Thus by Angle-Angle-Side criterion of congruence, we have

\(\Delta\)ABP \(\cong\) \(\Delta\)PCQ

The corresponding parts of the congruent triangles are equal

Therefore, AB = CQ ——– (1)

Now, DQ = DC + CQ

= DC + AB [from (1)]

Question 10:

In the given figure, OA = OB and OP = OQ. Prove that

(i) PX = QX

(ii) AX = BX

Solution:

Given: OA = OB and OP = OQ

To prove:

(i) PX = QX

(ii) AX = BX

Proof: In \(\Delta\)OAQ and \(\Delta\)OPB, we have,

OA = OB [Given]

\(\angle\)O = \(\angle\)O [Common]

OQ = OP [Given]

Thus by Side-Angle-Side criterion of congruence, we have

\(\Delta\)OAQ \(\cong\) \(\Delta\)OPB

The corresponding parts of the congruent triangles are equal.

Therefore, \(\angle\)OBP = \(\angle\)OAQ ——— (1)

Thus, in \(\Delta\)BXQ and \(\Delta\)PXA, we have

BQ = OB – OQ

and, PA = OA – OP

But, OP = OQ

and OA =OB [Given]

Therefore, we have, BQ = PA ———- (2)

Now consider triangles \(\Delta\)BXQ and \(\Delta\)PXA.

\(\angle\)BXQ = \(\angle\)PXA [vertically opposite angles]

\(\angle\)OBP = \(\angle\)OAQ [from (1)]

BQ = PA [from (2)]

Thus by Angle-Angle-Side criterion of congruence, we have,

Therefore, \(\Delta\)BXQ \(\cong\) \(\Delta\)PPXA

PX = QX [c.p.c.t]

AX = BX [c.p.c.t]’

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