NCERT Solutions for Class 9 Maths Chapter 9 - Congruence Of Triangles And Inequalities In A Triangle

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Question 1:

Question 2:

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

Question 3:

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.

Question 4:

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

Question 5:

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.

Question 6:

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

Question 7:

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.

Question 8:

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.

Question 9:

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.

Question 10:

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)

Question 11:

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

Hence, AC + AD = BC.

Question 12:

Proof:
$$\begin{gathered} \mathrm{In} △\mathrm{OQA} \mathrm{and} △\mathrm{OPB}, \mathrm{we} \mathrm{have}: \\ \mathrm{OQ}=\mathrm{OP} \left(\mathrm{Given}\right) \\ \mathrm{OA}=\mathrm{OB} \left(\mathrm{Given}\right) \\ \angle \mathrm{AOQ}=\angle \mathrm{BOP} \left(\mathrm{Common}\right) \\ △\mathrm{OQA}\cong \mathrm{OPB} (\mathrm{SAS} \mathrm{criterion}) \end{gathered}$$
$\angle OAQ=\angle OBP (\mathrm{Corresponding} \mathrm{angles} \mathrm{of} \mathrm{congruent} \mathrm{triangles})$
Now, consider triangles BQX and APX.
$$\begin{gathered} \mathrm{Given}: \\ \mathrm{OA}=\mathrm{OB} \\ \mathrm{OP}=\mathrm{OQ} \\ \therefore \mathrm{OA}-\mathrm{OP}=\mathrm{OB}-\mathrm{OQ} \\ \Rightarrow \mathrm{AP}=\mathrm{BQ} \end{gathered}$$
Further, $\angle \mathrm{BXQ}=\angle \mathrm{AXP} (\mathrm{Vertically} \mathrm{opposite} \mathrm{angles})$
Also, we have proven that $\angle QBX=\angle PAX$.
$∆\mathrm{BQX}\cong ∆\mathrm{APX} (\mathrm{AAS} \mathrm{criterion})$

$\therefore \mathrm{PX}=\mathrm{QX} (\mathrm{corresponding} \mathrm{sides} \mathrm{of} \mathrm{congruent} \mathrm{triangles})$
$\mathrm{Also}, \mathrm{AX}=\mathrm{BX} (\mathrm{corresponding} \mathrm{sides} \mathrm{of} \mathrm{congruent} \mathrm{triangles})$
Hence, proved.

Question 13:

$$\begin{gathered} \mathrm{Since} ∆\mathrm{ABC} \mathrm{is} \mathrm{an} \mathrm{equilateral} ∆, \mathrm{then} \\ \angle \mathrm{ABC} = \angle \mathrm{BCA} = \angle \mathrm{CAB} = 60° \\ \mathrm{Since} \mathrm{PQ}\parallel \mathrm{CA} \mathrm{and} \mathrm{PC} \mathrm{is} \mathrm{a} \mathrm{transversal}, \mathrm{then} \\ \angle \mathrm{BPQ}=\angle \mathrm{BCA}=60° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \mathrm{Since} \mathrm{PQ}\parallel \mathrm{CA} \mathrm{and} \mathrm{QA} \mathrm{is} \mathrm{a} \mathrm{transversal}, \mathrm{then} \\ \angle \mathrm{BQP}=\angle \mathrm{BAC}=60° \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \mathrm{Further}, \angle \mathrm{B}=60° \\ \mathrm{In} ∆\mathrm{BPQ}, \\ \angle \mathrm{B} = \angle \mathrm{BPQ} = \angle \mathrm{BQP} = 60° \\ \therefore △\mathrm{BPQ}\hspace{0.17em}\mathrm{is} \mathrm{an} \mathrm{equilateral} \mathrm{triangle}. \\ \mathrm{i}.\mathrm{e}., \mathrm{BP}=\mathrm{PQ}=\mathrm{BQ} \\ \mathrm{Now}, \mathrm{BP}=\mathrm{CR} \\ \Rightarrow \mathrm{PQ}=\mathrm{CR} ....\left(1\right) \\ \mathrm{Considering} △\mathrm{MPQ} \mathrm{and} △\mathrm{MCR}, \mathrm{we} \mathrm{get}: \\ \angle \mathrm{PQM}=\angle \mathrm{MRC} \left(\mathrm{Alternate} \mathrm{interior} \mathrm{angles}\right) \\ \angle \mathrm{PMQ}=\angle \mathrm{CMR} \left(\mathrm{Vertically} \mathrm{opposite} \mathrm{angles} \right) \\ \mathrm{PQ}=\mathrm{CR} \left[\mathrm{using} \left(1\right)\right] \\ ∆\mathrm{MPQ}\cong ∆\mathrm{MCR} \left(\mathrm{AAS} \mathrm{criterion}\right) \\ \Rightarrow \mathrm{MP}=\mathrm{MC} \left[\mathrm{Corresponding} \mathrm{parts} \mathrm{of} \mathrm{congruent} \mathrm{triangles} \mathrm{are} \mathrm{equal}\right] \\ \Rightarrow \mathrm{QR} \mathrm{bisects} \mathrm{PC}. \end{gathered}$$

Question 14:

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

∴ 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.

Question 15:

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

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

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

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

Question 16:

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.

Question 17:

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

Question 18:

$$\begin{gathered} \mathrm{In} △\mathrm{BDL} \mathrm{and} △\mathrm{CDM}, \mathrm{we} \mathrm{have}: \\ \mathrm{BD}=\mathrm{CD} (\mathrm{D} \mathrm{is} \mathrm{midpoint}) \\ \mathrm{DL}=\mathrm{DM} \left(\mathrm{Given}\right) \\ \angle \mathrm{BLD}=\angle \mathrm{CMD} (90° \mathrm{each}) \\ \mathrm{Thus},△\mathrm{BDL}\cong △\mathrm{CDM} (\mathrm{RHS} \mathrm{criterion}) \end{gathered}$$
$\Rightarrow \mathrm{BL}=\mathrm{MC} \left(\mathrm{CPCT}\right)$
∴$\angle B=\angle C$
This makes triangle ABC an isosceles triangle.
Or AB = AC
Hence, proved.

Question 19:

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.

Question 20:

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)

Question 21:

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)

Question 22:

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
$$\begin{gathered} \angle ABC+\angle BOC=\angle ABC+\angle DBA \\ \therefore \angle BOC=\angle DBA \end{gathered}$$

Question 23:

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.

Question 24:

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)

Question 25:

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.

Question 26:

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.

Question 27:

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.

Question 1:

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.

Question 2:

(i) No, because the sum of two sides of a triangle is not greater than the third side.
5 + 4 = 9

(ii) Yes, because the sum of two sides of a triangle is greater than the third side.
7 + 4 > 8; 8 + 7 > 4; 8 + 4 > 7

(iii) Yes, because the sum of two sides of a triangle is greater than the third side.
5 + 6 > 10; 10 + 6 > 5; 5 + 10 > 6

(iv) Yes, because the sum of two sides of a triangle is greater than the third side.
2.5 + 5 > 7; 5 + 7 > 2.5; 2.5 + 7 > 5

(v) No, because the sum of two sides of a triangle is not greater than the third side.
3 + 4 < 8

Question 3:

Given: In ΔABC, ∠A = 50° and ∠B = 60°

In ΔABC,
∠A + ∠B + ∠C = 180°           (Angle sum property of a triangle)
$\Rightarrow$50° + 60° + ∠C = 180°
$\Rightarrow$110° + ∠C = 180°
$\Rightarrow$∠C = 180° $-$ 110°
$\Rightarrow$∠C = 70°

Hence, the longest side will be opposite to the largest angle (∠C = 70°)  i.e. AB.
And, the shortest side will be opposite to the smallest angle (∠A = 50° ) i.e. BC.

Question 4:

(i) Given: In âˆ†ABC, ∠A = 90°

So, sum of the other two angles in triangle ∠B + ∠C = 90°

i.e. ∠B, ∠C < 90°

Since, ∠A is the greatest angle.

So, the longest side is BC.

(ii) Given: ∠A = ∠B = 45°

Using angle sum property of triangle,

∠C = 90°

Since, ∠C is the greatest angle.

So, the longest side is AB.

(iii) Given: ∠A = 100° and ∠C = 50°

Using angle sum property of triangle,

∠B = 30°

Since, ∠A is the greatest angle.

So, the shortest side is BC.