RS AGGARWAL CLASS 9 CHAPTER 9 CONGRUENCE OF TRIANGLES AND INEQUALITIES IN A TRIANGLE EXERCISE 9A

 EXERCISE 9A

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Q1 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 1:

In the given figure, AB || CD and O is the midpoint of AD.
Show that
(i) ΔAOB ≅ ΔDOC.
(ii) O is the midpoint of BC.

Answer 1:

Given : In the given figure, AB || CD and O is the midpoint of AD.

To prove:
(i) ΔAOB ≅ ΔDOC.
(ii) O is the midpoint of BC.

Proof:
(i) In ΔAOB and ΔDOC,
∠BAO = ∠CDO                    (Alternate interior angles, AB || CD)
AO = DO                              (Given, O is the midpoint of AD)
∠AOB = ∠DOC                   (Vertically opposite angles)

∴ By ASA congruence criteria,
ΔAOB ≅ ΔDOC

(ii) ∵ ΔAOB ≅ ΔDOC           [From (i)]
∴ BO = CO                           (CPCT)
Hence, O is the midpoint of BC.

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Question 2:

In the given figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Answer 2:

Given: In the given figure, AD and BC are equal perpendiculars to a line segment AB.

To prove: CD bisects AB

Proof:
In ΔAOD and ΔBOC,
∠DAO = ∠CBO = 90$°$           (Given)
AD = BC                              (Given)
∠DOA = ∠COB                    (Vertically opposite angles)

∴ By AAS congruence criteria,
ΔAOD ≅ ΔBOC
∴ AO = BO                           (CPCT)
Hence, CD bisects AB.

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Question 3:

In the given figure, two parallel line l and m are intersected by two parallel lines p and q.
Show that ΔABC ≅ ΔCDA.

Answer 3:

Given: In the given figure, two parallel line l and m are intersected by two parallel lines p and q.

To prove: ΔABC ≅ ΔCDA

Proof:
In ΔABC and ΔCDA,

$\angle$BAC = $\angle$DCA           (Alternate interior angles, p$\parallel$q)
$\angle$BCA = $\angle$DAC           (Alternate interior angles, l$\parallel$m)
AC = CA                        (Common side)

$\therefore$ By ASA congruence criteria,
ΔABC ≅ ΔCDA

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Question 4:

AD is an altitude of an isosceles ΔABC in which AB = AC.
Show that (i) AD bisects BC, (ii) AD bisects ∠A.

Answer 4:

Given: AD is an altitude of an isosceles ΔABC in which AB = AC.

To prove: (i) AD bisects BC, (ii) AD bisects ∠A

Proof:
(i) In ΔABD and ΔACD,

$\angle$ADB = $\angle$ADC = 90$°$          (Given, AD$\perp$BC)
AB = AC                                   (Given)
AD = AD                                  (Common side)

$\therefore$ By RHS congruence criteria,
ΔABD ≅ ΔACD

$\therefore$ BD = CD       (CPCT)

Hence, AD bisects BC.

(ii)
$\because$ ΔABD ≅ ΔACD       [From (i)]
$\therefore$ $\angle$BAD = $\angle$CAD      (CPCT)
Hence, AD bisects ∠A.

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Question 5:

In the given figure, BE and CF are two equal altitudes of ΔABC.
Show that (i) ΔABE ≅ ΔACF, (ii) AB = AC.

Answer 5:

$$\begin{gathered} \mathrm{In} △\mathrm{ABE} \mathrm{and} △\mathrm{ACF}, \mathrm{we} \mathrm{have}: \\ \mathrm{BE} =\mathrm{CF} \left(\mathrm{Given}\right) \\ \angle \mathrm{BEA}=\angle \mathrm{CFA}=90° \\ \angle \mathrm{A}=\angle \mathrm{A} \left(\mathrm{Common}\right) \\ △\mathrm{ABE}\cong △\mathrm{ACF} (\mathrm{AAS} \mathrm{criterion}) \end{gathered}$$
AB = AC (CPCT)

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Question 6:

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC.

Answer 6:

Given: ΔABC and ΔDBC are two isosceles triangles on the same base BC.

To prove:
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC

Proof:
(i) In ΔABD and ΔACD,

BD = CD                                  (Given, ΔDBC is an isosceles triangles)
AB = AC                                   (Given, ΔABC is an isosceles triangles)
AD = AD                                  (Common side)

$\therefore$ By SSS congruence criteria,
ΔABD ≅ ΔACD

Also, $\angle$BAD = $\angle$CAD         (CPCT)
or, $\angle$BAE = $\angle$CAE     .....(1)

(ii)
In ΔABE and ΔACE,

AB = AC                                   (Given, ΔABC is an isosceles triangles)
$\angle$BAE = $\angle$CAE                       [From (i)]
AE = AE                                  (Common side)

$\therefore$ By SAS congruence criteria,
ΔABE ≅ ΔACE

Also, BE = CE                    (CPCT)    .....(2)
And, $\angle$AEB = $\angle$AEC         (CPCT)   .....(3)

(iii)
In ΔBED and ΔCED,

BD = CD                                  (Given, ΔDBC is an isosceles triangles)
BE = CE                                   [From (2)]
DE = DE                                  (Common side)

$\therefore$ By SSS congruence criteria,
ΔBED ≅ ΔCED

Also, $\angle$BDE = $\angle$CDE         (CPCT)   .....(4)

(iv)
$\because$ $\angle$BAE = $\angle$CAE        [From (1)]
And, $\angle$BDE = $\angle$CDE   [From (4)]
$\therefore$ AE bisects ∠A as well as ∠D.

(v)
$$\begin{gathered} \because \angle AEB+\angle AEC=180° \left(\mathrm{Linear} \mathrm{pair}\right) \\ \Rightarrow \angle AEB+\angle AEB=180° \left[\mathrm{From} \left(3\right)\right] \\ \Rightarrow 2\angle AEB=180° \\ \Rightarrow \angle AEB=\frac{180°}{2} \\ \Rightarrow \angle AEB=90° .....\left(5\right) \end{gathered}$$

From (2) and (5), we get
AE is the perpendicular bisector of BC.

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Question 7:

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

Answer 7:

Consider the triangles AEB and CDB.

$\angle EBA=\angle DBC$  (Common angle)  ...(i)

Further, we have:

$$\begin{gathered} \angle \mathrm{BEA}=180-\mathrm{y} \\ \angle \mathrm{BDC}=180-\mathrm{x} \\ \mathrm{Since} \mathrm{x} = \mathrm{y}, \mathrm{we} \mathrm{have}: \\ 180-\mathrm{x} =180-\mathrm{y} \\ \Rightarrow \angle \mathrm{BEA}=\angle \mathrm{BDC} ...\left(\mathrm{ii}\right) \end{gathered}$$

AB = CB     (Given) ...(iii)
From (i), (ii) and (iii), we have:
$△BDC\cong △BEA$   (AAS criterion)
∴ AE = CD (CPCT)
Hence, proved.

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Q8 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 8:

In the given figure, line l is the bisector of an angle ∠A and B is any point on l. If BP and BQ are perpendiculars from B to the arms of ∠A, show that
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Answer 8:

Given: In the given figure, ∠BAQ = ∠BAP, BP$\perp$AP and BQ$\perp$AQ.

To prove:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ, i.e., B is equidistant from the arms of ∠A.

Proof:
(i) In ΔAPB and ΔAQB,

∠BAQ = ∠BAP,                   (Given)
∠APB = ∠AQB = 90$°$          (Given, BP$\perp$AP and BQ$\perp$AQ)
AB = AB                               (Common side)

$\therefore$ By AAS congruence criteria,
ΔAPB ≅ ΔAQB

(ii)
$\because$ ΔAPB ≅ ΔAQB            [From (i)]
$\therefore$ BP = BQ                       (CPCT)
Hence, B is equidistant from the arms of ∠A.

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Question 9:

ABCD is a quadrilateral such that diagonal AC bisects the angles ∠A and ∠C. Prove that AB = AD and CB = CD.

Answer 9:

Given: In quadrilateral ABCD, AC bisects the angles ∠A and ∠C.

To prove: AB = AD and CB = CD

Proof:
In $∆$ABC and $∆$ADC,

$\angle$BAC = $\angle$DAC            (Given, AC bisects the angles ∠A)
AC = AC                        (Common side)
$\angle$BCA = $\angle$DCA            (Given, AC bisects the angles ∠C)

$\therefore$ By ASA congruence criteria,
$∆$ABC $\cong$$∆$ADC

Hence, AB = AD and CB = CD.     (CPCT)

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Question 10:

ΔABC is a right triangle right angled at A such that AB = AC and bisector of ∠C intersects the side AB at D. Prove that AC + AD = BC.

Answer 10:

Given: In right triangle ΔABC, ∠BAC = 90$°$, AB = AC and ∠ACD = ∠BCD. 

To prove: AC + AD = BC

Proof:
Let AB = AC = x and AD = y.

In $∆$ABC,

$$\begin{gathered} B{C}^2=A{B}^2+A{C}^2 \\ \Rightarrow B{C}^2=x^2+x^2 \\ \Rightarrow B{C}^2=2x^2 \\ \Rightarrow BC=\sqrt{2x^2} \\ \Rightarrow BC=x\sqrt{2} \end{gathered}$$

Now,

$\frac{BD}{AD}=\frac{BC}{AC}$                  (An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.)
$\Rightarrow \frac{x-y}{y}=\frac{x\sqrt{2}}{x}$

$\Rightarrow \frac{x}{y}-\frac{y}{y}=\sqrt{2}$

$$\begin{gathered} \Rightarrow \frac{x}{y}-1=\sqrt{2} \\ \Rightarrow \frac{x}{y}=\left(\sqrt{2}+1\right) \\ \Rightarrow y=\frac{x}{\left(\sqrt{2}+1\right)} \end{gathered}$$
$$\begin{gathered} \Rightarrow y=\frac{x}{\left(\sqrt{2}+1\right)}\times \frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)} \\ \Rightarrow y=\frac{x\left(\sqrt{2}-1\right)}{{\left(\sqrt{2}\right)}^2-1^2} \\ \Rightarrow y=\frac{x\sqrt{2}-x}{2-1} \\ \Rightarrow y=x\sqrt{2}-x \\ \Rightarrow x+y=x\sqrt{2} \end{gathered}$$

Hence, AC + AD = BC.

PAGE NO-284

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Question 11:

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX, (ii) AX = BX.

Answer 11:


Proof:
In △OQA and △OPB, we have :

OQ=OP (Given)

OA=OB (Given)

∠AOQ=∠BOP (Common)

△OQA≅OPB (SAS criterion)
∠OAQ=∠OBP (Corresponding angles of congruent triangles)
Now, consider triangles BQX and APX.

Given:

OA=OB

OP=OQ

∴ OA−OP=OB−OQ

⇒AP=BQ
Further, ∠BXQ=∠AXP (Vertically opposite angles)
Also, we have proven that ∠QBX=∠PAX.
ΔBQX≅ΔAPX (AAS criterion)

∴ PX=QX (corresponding sides of congruent triangles)
Also, AX=BX (corresponding sides of congruent triangles)
Hence, proved.

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Question 12:

In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

Answer 12:

Since ΔABC is an equilateral Δ, then ∠ABC = ∠BCA = ∠CAB = 60°

Since PQ∥CA and PC is a transversal, then ∠BPQ=∠BCA=60° (Corresponding angles)

Since PQ∥CA and QA is a transversal, then ∠BQP=∠BAC=60° (Corresponding angles)

Further, ∠B=60°In ΔBPQ, ∠B = ∠BPQ = ∠BQP = 60°

∴△BPQ is an equilateral triangle .i.e., BP=PQ=BQ

Now, BP=CR⇒PQ=CR ....(1)

Considering △MPQ and △MCR, we get:∠PQM=∠MRC (Alternate interior angles)

∠PMQ=∠CMR (Vertically opposite angles ) 

 PQ=CR [using (1)] 

ΔMPQ≅ΔMCR (AAS criterion)

⇒MP=MC [Corresponding parts of congruent triangles are equal]

⇒QR bisects PC.

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Question 13:

In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint of BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

Answer 13:

$$\begin{gathered} \mathrm{In} △\mathrm{ABP} \mathrm{and} △\mathrm{QCP}, \mathrm{we} \mathrm{have}: \\ \angle \mathrm{BPA}=\angle \mathrm{CPQ} (\mathrm{Vertically} \mathrm{opposite} \mathrm{angle}) \\ \angle \mathrm{PAB}=\angle \mathrm{PQC} (\mathrm{Alternate} \mathrm{angles}) \\ \mathrm{PB} = \mathrm{PC} (\mathrm{P} \mathrm{is} \mathrm{the} \mathrm{midpoint}) \\ △\mathrm{ABP}\cong △\mathrm{QCP} (\mathrm{AAS} \mathrm{criterion}) \end{gathered}$$

∴ AB = CQ            (CPCT)

$$\begin{gathered} \mathrm{DQ}=\mathrm{DC}+\mathrm{CQ} \\ \Rightarrow \mathrm{DQ}=\mathrm{DC}+\mathrm{AB} (\mathrm{Proved}, \mathrm{AB} = \mathrm{CQ}) \end{gathered}$$
Hence, proved.

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Question 14:

In the given figure, ABCD is a square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

Answer 14:

$$\begin{gathered} \mathrm{In} △\mathrm{PAD} \mathrm{and} △\mathrm{PAB}, \mathrm{we} \mathrm{have}: \\ \mathrm{AD}=\mathrm{AB} (\mathrm{Side} \mathrm{of} \mathrm{a} \mathrm{square}) \\ \mathrm{AP}=\mathrm{AP} \left(\mathrm{Common}\right) \\ \mathrm{PD} =\mathrm{PB} \left(\mathrm{Given}\right) \\ △\mathrm{PAD}\cong \mathrm{PAB} (\mathrm{SSS} \mathrm{criterion}) \\ \Rightarrow \angle \mathrm{APD}=\angle \mathrm{APB} \end{gathered}$$

$$\begin{gathered} \mathrm{In} △\mathrm{CPD} \mathrm{and} △\mathrm{CPB}, \mathrm{we} \mathrm{have}: \\ \mathrm{CD} = \mathrm{CB} (\mathrm{Sides} \mathrm{of} \mathrm{square}) \\ \mathrm{CP}=\mathrm{CP} \left(\mathrm{Common}\right) \\ \mathrm{PD}=\mathrm{PB} \left(\mathrm{Given}\right) \\ △\mathrm{CPD}\cong △\mathrm{CPB} (\mathrm{SSS} \mathrm{test}) \\ \Rightarrow \angle \mathrm{CPD}=\angle \mathrm{CPB} \end{gathered}$$

$$\begin{gathered} \therefore \angle \mathrm{APD}+\angle \mathrm{CPD}=\angle \mathrm{APB}+\angle \mathrm{CPB} \\ \mathrm{But}\angle \mathrm{APD}+\angle \mathrm{CPD}+\angle \mathrm{APB}+\angle \mathrm{CPB}=360° \\ \Rightarrow \angle \mathrm{APD}+\angle \mathrm{CPD}=180° \end{gathered}$$

So, CPA is a straight line.
Hence, proved.

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Question 15:

In the given figure, O is a point in the interior of square ABCD such that ΔOAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.

Answer 15:

Given: In square ABCD, ΔOAB is an equilateral triangle.

To prove: ΔOCD is an isosceles triangle.

Proof:

$$\begin{gathered} \because \angle DAB=\angle CBA=90° \left(\mathrm{Angles} \mathrm{of} \mathrm{square} ABCD\right) \\ \mathrm{And}, \angle OAB=OBA=60° \left(\mathrm{Angles} \mathrm{of} \mathrm{equilateral} ∆OAB\right) \\ \therefore \angle DAB-\angle OAB=\angle CBA-\angle OBA=90°-60° \\ \Rightarrow \angle OAD=\angle OBC=30° .....\left(\mathrm{i}\right) \end{gathered}$$

Now, in ΔDAO and ΔCBO,

AD = BC                   (Sides of square ABCD)
$\angle$DAO = $\angle$CBO      [From (i)]
AO = BO                  (Sides of equilateral ΔOAB)

$\therefore$ By SAS congruence criteria,
ΔDAO $\cong$ ΔCBO

So, OD = OC         (CPCT)
Hence, ΔOCD is an isosceles triangle.

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Question 16:

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ∆ABC such that AX = AY. Prove that CX = BY.

Answer 16:

$$\begin{gathered} \mathrm{In} △\mathrm{AXC} \mathrm{and} △\mathrm{AYB}, \mathrm{we} \mathrm{have}: \\ \mathrm{AC}=\mathrm{AB} \left(\mathrm{Given}\right) \\ \mathrm{AX}=\mathrm{AY} \left(\mathrm{Given}\right) \\ \angle \mathrm{BAC}=\angle \mathrm{CAB} (\mathrm{Angle} \mathrm{common} \mathrm{to} △\mathrm{AXC} \mathrm{and} △\mathrm{AYB}) \\ \therefore △\mathrm{AXC}\cong △\mathrm{AYB} (\mathrm{SAS} \mathrm{criterion}) \\ \mathrm{So}, \mathrm{CX}=\mathrm{BY} \left(\mathrm{CPCT}\right) \end{gathered}$$
Hence, proved.

PAGE NO-285

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Q17 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 17:

In ∆ABC, D is the midpoint of BC. If DL ⊥ AB and DM ⊥ AC such that DL = DM, prove that AB = AC.

Answer 17:

In △BDL and △CDM, we have:

BD=CD (D is midpoint)

DL=DM (Given)

∠BLD=∠CMD (90° each)

Thus, △BDL≅△CDM (RHS criterion)

⇒BL=BC (C.P.C.T)

∴∠B=∠C

This makes triangle ABC an isosceles triangle.
Or AB = AC
Hence, proved.

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Q18 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 18:

In ∆ABC, AB = AC and the bisectors of ∠B and ∠C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠A.

Answer 18:

In triangle ABC, we have:
AB = AC   (Given)
$$\begin{gathered} \Rightarrow \angle B=\angle C \\ \Rightarrow \frac{1}{2}\angle B=\frac{1}{2}\angle C \\ \Rightarrow \angle OBC=\angle OCB \\ \Rightarrow BO=CO \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \mathrm{in} △AOB \mathrm{and} △AOC, \mathrm{we} \mathrm{have}: \\ OB=OC \left(\mathrm{Proved}\right) \\ AB=AC \left(\mathrm{Given}\right) \\ AO=AO \left(\mathrm{Common}\right) \\ \therefore △AOB\cong △AOC (\mathrm{SSS} \mathrm{criterion}) \end{gathered}$$
i.e., $\angle BAO=\angle CAO (\mathrm{Corresponding} \mathrm{angles} \mathrm{of} \mathrm{congruent} \mathrm{triangles})$

So, it shows that ray AO is the bisector of $\angle A$.
Hence, proved.

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Question 19:

The line segments joining the midpoints M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.

Answer 19:

Given: In trapezium ABCD, M and N are mid-points of AB and DC, MN$\perp$AB and MN$\perp$DC.

To prove: AD = BC

Construction: Join CM and DM.

Proof:

In ΔCMN and ΔDMN,

MN = MN                           (Common sides)
$\angle$CNM = $\angle$DNM = 90$°$     (Given, MN$\perp$DC)
CN = DN                             (Given, N is the mid-point DC)

$\therefore$ By SAS congruence criteria,
ΔCMN $\cong$ ΔDMN

So, CM = DM                  (CPCT)    .....(i)
And, $\angle$CMN = $\angle$DMN    (CPCT)
But, $\angle$AMN = $\angle$BMN = 90$°$      (Given, MN$\perp$AB)
$\Rightarrow$$\angle$AMN $-$ $\angle$CMN = $\angle$BMN $-$ $\angle$DMN
$\Rightarrow$$\angle$AMD = $\angle$BMC           .....(ii)

Now, in ΔAMD and ΔBMC,

DM = CM                            [From (i)]
$\angle$AMD = $\angle$BMC                [From (ii)]
AM = BM                            (Given, M is the mid-point AB)

$\therefore$ By SAS congruence criteria,
ΔAMD $\cong$ ΔBMC

Hence, AD = BC    (CPCT)

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Q20 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 20:

The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that ∠MOC = ∠ABC.

Answer 20:

Given: In isosceles $∆$ABC, AB = AC; OB and OC are bisectors of ∠B and ∠C, respectively.

To prove: ∠MOC = ∠ABC

Proof:

In ∆ABC,

$\because$AB = AC             (Given)
$\therefore$∠ABC = ∠ACB   (Angles opposite to equal sides are equal)
$\Rightarrow \frac{1}{2}$∠ABC = $\frac{1}{2}$∠ACB
$\Rightarrow$∠OBC = ∠OCB         (Given, OB and OC are the bisectors of ∠B and ∠C, respectively)    .....(i)

Now, in ∆OBC, ∠MOC is an exterior angle
$\Rightarrow$∠MOC = ∠OBC + ∠OCB    (An exterior angle is equal to the sum of two opposite interior angles)
$\Rightarrow$∠MOC = ∠OBC + ∠OBC    [From (i)]
$\Rightarrow$∠MOC = 2∠OBC
Hence, ∠MOC = ∠ABC    (Given, OB is the bisector of ∠B)

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Q21 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 21:

The bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC intersect each other at a point O. Show that the exterior angle adjacent to ∠ABC is equal to ∠BOC.

Answer 21:

Given: In an isosceles ΔABC, AB = AC, BO and CO are the bisectors of ∠ABC and ∠ACB, respectively.

To prove: ∠ABD = ∠BOC

Construction: Produce CB to point D.

Proof:

In ΔABC,

$\because$ AB = AC                    (Given)
$\therefore$ ∠ACB = ∠ABC            (Angle opposite to equal sides are equal)

$$\begin{gathered} \Rightarrow \frac{1}{2}\angle ACB=\frac{1}{2}\angle ABC \\ \Rightarrow \angle OCB=\angle OBC .....\left(\mathrm{i}\right) \\ \left(\mathrm{Given}, BO \mathrm{and} CO \mathrm{are} \mathrm{angle} \mathrm{bisector} \mathrm{of} \angle ABC \mathrm{and} \angle ACB, \mathrm{respectively}\right) \end{gathered}$$

In ΔBOC,

$$\begin{gathered} \angle OBC+\angle OCB+\angle BOC=180° \left(\mathrm{By} \mathrm{angle} \mathrm{sum} \mathrm{property} \mathrm{of} \mathrm{triangle}\right) \\ \Rightarrow \angle OBC+\angle OBC+\angle BOC=180° \left[\mathrm{From} \left(\mathrm{i}\right)\right] \\ \Rightarrow 2\angle OBC+\angle BOC=180° \\ \Rightarrow \angle ABC+\angle BOC=180° \left(BO \mathrm{is} \mathrm{the} \mathrm{angle} \mathrm{bisector} \mathrm{of} \angle ABC\right) .....\left(\mathrm{ii}\right) \end{gathered}$$

Also, DBC is a straight line.
So, $\angle ABC+\angle DBA=180° \left(\mathrm{Linear} \mathrm{pair}\right) .....\left(\mathrm{iii}\right)$

From (ii) and (iii), we get
$$\begin{gathered} \angle ABC+\angle BOC=\angle ABC+\angle DBA \\ \therefore \angle BOC=\angle DBA \end{gathered}$$

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Q22 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 22:

P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meets BC at Q, prove that ΔBPQ is an isosceles triangle.

Answer 22:

Given: BP is the bisector of ∠ABC, and BA$\parallel$QP

To prove: ΔBPQ is an isosceles triangle

Proof:

$$\begin{gathered} \because \angle 1=\angle 2 \left(\mathrm{Given}, BP \mathrm{is} \mathrm{the} \mathrm{bisector} \mathrm{of} \angle ABC\right) \\ \mathrm{And}, \angle 1=\angle 3 \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right) \\ \therefore \angle 2=\angle 3 \\ \mathrm{So}, PQ=BQ \left(\mathrm{In} \mathrm{a} \mathrm{triangle}, \mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{are} \mathrm{equal}.\right) \end{gathered}$$

But these are sides of $∆BPQ$.

Hence, $∆BPQ$ is an isosceles triangle.

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Q23 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 23:

The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D, as shown in the figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.

Answer 23:

Given: An object is placed at a point A, the image of the object is seen at the point B, an observer is at point D, and LM is a plane mirror.

To Prove: The image is as far behind the mirror as the object is in front of the mirror, i.e. BT = AT.

Proof:

$\because$ LM is a plane mirror

$\therefore$ $i = r$              (Angle of incidence is always equal to angle of reflection)    .....(1)

Also, $AB\parallel CN$                            (Both AB and CN are perpendicular to LM)

$\Rightarrow \angle TAC=\angle ACN=i$                (Alternate interior angles)      .....(2)

And, $\angle CBT=\angle NCD=r$           (Corresponding angles)         .....(3)

From (1), (2) and (3), we get

$\angle TAC = \angle CBT$        .....(4)

Now,

$$\begin{gathered} \mathrm{In} ∆TAC \mathrm{and} ∆CBT, \\ \angle TAC=\angle CBT \left[\mathrm{From} \left(4\right)\right] \\ \angle ATC=\angle BTC=90° \\ CT=CT \left(\mathrm{Common} \mathrm{side}\right) \\ \therefore \mathrm{By} \mathrm{AAS} \mathrm{congruence} \mathrm{criteria}, \\ ∆TAC\cong ∆CBT \end{gathered}$$

Hence, AT = BT    (CPCT)

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Q24 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 24:

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

Answer 24:

Let AB be the breadth of the river.
M is any point situated on the bank of the river.
Let O be the mid point of BM.
​Moving along perpendicular to point such that A,O and N are in a straight line.
Then MN is the required breadth of the river.

$$\begin{gathered} \mathrm{In} △\mathrm{OBA} \mathrm{and} △\mathrm{OMN}, \mathrm{we} \mathrm{have}: \\ \mathrm{OB}=\mathrm{OM} (\mathrm{O} \mathrm{is} \mathrm{midpoint}) \\ \angle \mathrm{OBA}=\angle \mathrm{OMN} (\mathrm{Each} 90°) \\ \angle \mathrm{AOB}=\angle \mathrm{NOM} (\mathrm{Vertically} \mathrm{opposite} \mathrm{angle}) \\ \therefore △\mathrm{OBA}\cong △\mathrm{OMN} (\mathrm{ASA} \mathrm{criterion}) \end{gathered}$$

Thus, MN = AB (CPCT)
If MN is known, one can measure the width of the river without actually crossing it.

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Q25 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 25:

In a ΔABC, D is the midpoint of side AC such that BD = $\frac{1}{2}$AC. Show that ∠ABC is a right angle.

Answer 25:

Given: In $∆$ABC, D is the midpoint of side AC such that BD = $\frac{1}{2}$AC.

To prove: ∠ABC is a right angle.

Proof:

$$\begin{gathered} \mathrm{In} ∆ADB, \\ AD=BD \left(\mathrm{Given}, BD = \frac{1}{2}AC\right) \\ \Rightarrow \angle DAB=\angle DBA=x \left(\mathrm{Let}\right) \left(\mathrm{Angles} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{are} \mathrm{equal}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Similarly}, \mathrm{in} ∆DCB, \\ BD=CD \left(\mathrm{Given}\right) \\ \Rightarrow \angle DBC=\angle DCB=y \left(\mathrm{Let}\right) \end{gathered}$$

$\mathrm{In} ∆ABC,$

$$\begin{gathered} \angle ABC+\angle BCA+\angle CAB=180° \left(\mathrm{Angle} \mathrm{sum} \mathrm{property}\right) \\ \Rightarrow x+x+y+y=180° \\ \Rightarrow 2(x+y)=180° \\ \Rightarrow x+y=90° \\ \Rightarrow \angle ABC=90° \end{gathered}$$

Hence, ∠ABC is a right angle.

PAGE NO-286

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Q26 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 26:

“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle then the two triangles must be congruent.”
Is the statement true? Why?

Answer 26:

No, the statement is not true because the two triangles are congruent only by SAS congruence condition but the statement contains ASS or SSA condition as well which are not any condition for congruence of triangles.

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Q27 | Ex-9A |Class 9 | RS AGGARWAL | CONGRUENCE OF TRIANGLES AND INEQUALITIES IN  A TRIANGLE |Ch-9 |myhelper

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Question 27:

“If two angles and a side of one triangle are equal to two angles and a side of another triangle then the two triangles must be congruent.” Is the statement true? Why?

Answer 27:

Yes, the statement is true because the two triangles can be congruent by either AAS or ASA congruence criteria.

Disclaimer: If corresponding angles of two triangles are equal, then the by angle sum property, the third corresponding angle will be equal. So, if we have two corresponding angles and a corresponding side are equal, then the triangles can be proved congruent by SAS congruence criteria. Therefore, ASA and SAS criteria are treated as same.