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#### Page No 282: 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.

#### Page No 282: 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)
DOA = ∠COB                    (Vertically opposite angles)

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

#### Page No 282: 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

#### Page No 283: Given: AD is an altitude of an isosceles ΔABC in which AB AC.

Proof:
(i) In ΔABD and ΔACD,

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

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

$\therefore$ BD = CD       (CPCT)

(ii)
$\because$ ΔABD ≅ ΔACD       [From (i)]
$\therefore$ $\angle$BAD = $\angle$CAD      (CPCT)

AB = AC (CPCT)

#### Page No 283: 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)

$\therefore$ 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)
BECE                                   [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)

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

#### Page No 283:

Consider the triangles AEB and CDB.

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

Further, we have:

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

#### Page No 283: Given: In the given figure, ∠BAQ = ∠BAPBP$\perp$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, is equidistant from the arms of ∠A.

#### Page No 283: Given: In quadrilateral ABCDAC 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)

#### Page No 283: Given: In right triangle ΔABC, ∠BAC = 90$°$AB AC and ∠ACD = ∠BCD

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

In $∆$ABC,

$B{C}^{2}=A{B}^{2}+A{C}^{2}\phantom{\rule{0ex}{0ex}}⇒B{C}^{2}={x}^{2}+{x}^{2}\phantom{\rule{0ex}{0ex}}⇒B{C}^{2}=2{x}^{2}\phantom{\rule{0ex}{0ex}}⇒BC=\sqrt{2{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒BC=x\sqrt{2}$

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.)
$⇒\frac{x-y}{y}=\frac{x\sqrt{2}}{x}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{y}-\frac{y}{y}=\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{y}-1=\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{y}=\left(\sqrt{2}+1\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{x}{\left(\sqrt{2}+1\right)}$
$⇒y=\frac{x}{\left(\sqrt{2}+1\right)}×\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}\phantom{\rule{0ex}{0ex}}⇒y=\frac{x\left(\sqrt{2}-1\right)}{{\left(\sqrt{2}\right)}^{2}-{1}^{2}}\phantom{\rule{0ex}{0ex}}⇒y=\frac{x\sqrt{2}-x}{2-1}\phantom{\rule{0ex}{0ex}}⇒y=x\sqrt{2}-x\phantom{\rule{0ex}{0ex}}⇒x+y=x\sqrt{2}$

Hence, AC + AD = BC.

#### Page No 284:

Proof:

Now, consider triangles BQX and APX.

Further,
Also, we have proven that $\angle QBX=\angle PAX$.

Hence, proved.

#### Page No 284:

∴ AB = CQ            (CPCT)

Hence, proved.

#### Page No 284:

$\therefore \angle \mathrm{APD}+\angle \mathrm{CPD}=\angle \mathrm{APB}+\angle \mathrm{CPB}\phantom{\rule{0ex}{0ex}}\mathrm{But}\angle \mathrm{APD}+\angle \mathrm{CPD}+\angle \mathrm{APB}+\angle \mathrm{CPB}=360°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{APD}+\angle \mathrm{CPD}=180°$

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

#### Page No 284: Given: In square ABCD, ΔOAB is an equilateral triangle.

To prove: ΔOCD is an isosceles triangle.

Proof:

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.

Hence, proved.

#### Page No 285:

$\angle B=\angle C$
This makes triangle ABC an isosceles triangle.
Or AB = AC
Hence, proved.

#### Page No 285:

In triangle ABC, we have:
AB = AC   (Given)
$⇒\angle B=\angle C\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\angle B=\frac{1}{2}\angle C\phantom{\rule{0ex}{0ex}}⇒\angle OBC=\angle OCB\phantom{\rule{0ex}{0ex}}⇒BO=CO$

i.e.,

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

#### Page No 285: Given: In trapezium ABCDand are mid-points of AB and DC, MN$\perp$AB and MN$\perp$DC.

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)
$⇒$$\angle$AMN $-$ $\angle$CMN$\angle$BMN $-$ $\angle$DMN
$⇒$$\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

#### Page No 285: 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)
$⇒\frac{1}{2}$ABC = $\frac{1}{2}$ACB
$⇒$OBC = ∠OCB         (Given, OB and OC are the bisectors of ∠B and ∠C, respectively)    .....(i)

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

#### Page No 285: 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)

In ΔBOC,

Also, DBC is a straight line.
So,

From (ii) and (iii), we get

#### Page No 285: Given: BP is the bisector of ∠ABC, and BA$\parallel$QP

To prove: ΔBPQ is an isosceles triangle

Proof:

But these are sides of $∆BPQ$.

Hence, $∆BPQ$ is an isosceles triangle.

#### Page No 285: 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

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

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

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

And,            (Corresponding angles)         .....(3)

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

.....(4)

Now,

Hence, AT = BT    (CPCT)

#### Page No 285:

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.

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

#### Page No 285: Given: In $∆$ABCD is the midpoint of side AC such that BD $\frac{1}{2}$AC.

To prove: ∠ABC is a right angle.

Proof:

Hence, ∠ABC is a right angle.

#### Page No 286:

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.

#### Page No 286:

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.

#### Page No 296:

(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

#### Page No 296:

Given: In ΔABC, ∠A = 50° and ∠B = 60°

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

#### Page No 296:

(i) Given: In ABCA = 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.

#### Page No 297:

In triangle CBA, CBD is an exterior angle.

Triangle BCD is isosceles and BC = BD.
Let .

#### Page No 297: Given: ∠B < ∠and ∠C < ∠D

Proof:

#### Page No 297:

Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.

To prove:
​(i) ∠A > ∠C
(ii) ∠> ∠D Construction: Join AC.

Proof:

(ii) Construction: Join BD.

Proof:

#### Page No 297:

To prove: (AB + BC + CD + DA) > (AC + BD)

Proof: Adding (i), (ii), (iii) and (iv), we get

2(AB + BC + CD + DA) < 2( AC + BD)

Hence, (AB + BC + CD + DA) < (AC + BD).

#### Page No 297:

To prove: (AB + BC + CD + DA) < 2(BD + AC).
Proof: In ∆AOB,

In ∆BOC,

In ∆COD,

In ∆AOD,

#### Page No 297: Given: In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.

#### Page No 297:

Since the angle opposite to the longer side is greater, we have:

$PQ>PR\phantom{\rule{0ex}{0ex}}⇒\angle R>\angle Q\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\angle R>\frac{1}{2}\angle Q\phantom{\rule{0ex}{0ex}}⇒\angle SRQ>\angle RQS\phantom{\rule{0ex}{0ex}}⇒QS>SR$

$SQ>SR$

#### Page No 297: Given: In $∆$ABC, AB = AC

To prove: CD BD

Proof:

In $∆$ABC,

Since, AB = AC       (Given)

So, $\angle ABC=\angle ACB$      ...(i)

#### Page No 297: Given: In $∆$ABC, BC is the longest side.

To prove: $\angle$BAC > $\frac{2}{3}$ of a right angle, i.e., $\angle$BAC > 60$°$

Construct: Mark a point D on side AC such that AD = AB = BD.

Proof:

In $∆$ABD,

$\because$ AD = AB = BD    (By construction)

$\therefore \angle 1=\angle 3=\angle 4=60°$

Hence, $\angle$BAC > $\frac{2}{3}$ of a right angle.

#### Page No 298: To prove:
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC

Proof:
(i)

(ii)

#### Page No 298: Given:
In triangle ABC, O is any interior point.
We know that any segment from a point O inside a triangle to any vertex of the triangle cannot be longer than the two sides adjacent to the vertex.
Thus, OA cannot be longer than both AB and CA (if this is possible, then O is outside the triangle).

(i) OA cannot be longer than both AB and CA.​

Adding the above three equations, we get:
...(6)

OA cannot be longer than both AB and CA.​

Thus, the first equation to be proved is shown correct.

(iii) Now, consider the triangles OAC, OBA and OBC.
We have:

#### Page No 298: Given: AD ⊥ BC and CD BD

To prove: AC AB

Proof:

In $∆ABD$,

Using angle sum property of a triangle,

In $∆ADC$,

Using angle sum property of a triangle,

From (2), (3) and (4), we get

$\angle B>\angle C$

Therefore, $AC>AB$.

#### Page No 298: Given: CD DE

To prove: AB AC BE

Proof:

In $∆ABC$,

In $∆BED$,

From (1) and (2), we get

AB AC BE

#### Page No 300:

(a) SSA
SSA is not a criterion for congruence of triangles.

#### Page No 300:

(c) $∆CAB\cong ∆PQR$
As,
AB = QR,   (given)
BC = RP,     (given)
CA = PQ     (given)

$∆CAB\cong ∆PQR$

#### Page No 301:

(a) BC = PQ

If $∆ABC\cong ∆PQR$, then
BC = QR
Hence, the correct answer is option (a).

#### Page No 301:

In $△ABC$, we have:
AB = AC
B = 50°
Since ABC is an isosceles triangle, we have:
$\angle C=\angle B$
$\angle C=50°$
In triangle ABC, we have:
$\angle A+\angle B+\angle C=180°\phantom{\rule{0ex}{0ex}}⇒\angle A+50+50=180°\phantom{\rule{0ex}{0ex}}⇒\angle A=180°-100°\phantom{\rule{0ex}{0ex}}\therefore \angle A=80°$
Hence, the correct answer is option (c).

#### Page No 301:

Given: In ABCBC AB and ∠B = 80°.

In ABC,

As, $AB=BC$

$⇒\angle A=\angle C$

Let $\angle A=\angle C=x$

Using angle sum property of a triangle,

$⇒x=\frac{100°}{2}\phantom{\rule{0ex}{0ex}}⇒x=50°\phantom{\rule{0ex}{0ex}}⇒\angle A=50°$

Hence, the correct option is (a).

#### Page No 301: Given: In ABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm.

In ABC,

As, ∠C = ∠A                   (Given)

Therefore, $BC=AB$       (Sides opposite to equal angles.)

Hence, the correct option is (a).

#### Page No 301:

Since, 4 + 2.5 = 6.5

So, 6.5 cm cannot be the third side of the triangle, as the sum of two sides of a triangle is always greater than the third side.

Hence, the correct option is (b).

#### Page No 301:

(b) AB > AC

In $∆ABC$, we have:
$\angle C>\angle B$
The side opposite to the greater angle is larger.
$\therefore AB>AC$

#### Page No 301:

(b)​ $\angle E=60°$

$∆ABC\cong ∆FDE$
AB = 5cm, $\angle B=40°,\angle A=80°$ and FD = 5cm

$\angle E=60°$

#### Page No 301:

(c) AB

In triangle ABC, we have:
$\angle A=40°,\angle B=60°$        ...(Given)

∴ The side opposite to$\angle C$, i.e., AB, is the longest side of triangle ABC.

#### Page No 301:

$AB>AC$ is given.

$\angle ACB>\angle ABC$

#### Page No 301:

(b) OB > OC

AB >AC    (Given)
$⇒\angle C>\angle B$
$⇒\frac{1}{2}\angle C>\frac{1}{2}\angle B$
$⇒\angle OCB>\angle OBC$  (Given)
$⇒OB>OC$

#### Page No 301:

(a) 1:1

​In

i.e., $\angle ABO=\angle ACO$
∴

#### Page No 302:

(b) isosceles In

To prove: AB = AC

#### Page No 302:

(b) $\angle B=\angle E$ In
AB = DE      (Given)
BC = EF       (Given)
In order that $△ABC\cong DEF$, we must have $\angle B=\angle E$.

#### Page No 302: In order that $△ABC\cong △DEF$, we must have BC = EF.
Hence, the correct answer is option (c).

#### Page No 302:

(a) isosceles but not congruent

Thus, both the triangles are isosceles but not congruent.

#### Page No 302:

(c) A triangle can have two acute angles.
The sum of two acute angles is always less than 180o, thus satisfying the angle sum property of a triangle.
Therefore, a triangle can have two acute angles.

#### Page No 302:

a) Sum of any two sides of a triangle > the third side

b) Difference of any two sides of a triangle < the third side

c) Sum of three altitudes of a triangle < sum of its three side

d) Sum of any two sides of a triangle > twice the median to the 3rd side

e) Perimeter of a triangle > sum of its three medians

#### Page No 302:

a) Each angle of an equilateral triangle measures $60°$.
d) Drawing a $△ABC$ with AB = 3cm, BC = 4cm and CA = 7cm is not possible.