Rs Aggarwal 2019 2020 Solutions for Class 9 Math Chapter 9 Congruence Of Triangles And Inequalities In A Triangle are provided here with simple step-by-step explanations. These solutions for Congruence Of Triangles And Inequalities In A Triangle are extremely popular among Class 9 students for Math Congruence Of Triangles And Inequalities In A Triangle Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 9 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### 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.  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.

#### Question 2:

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

#### 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.  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 4:

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

#### Question 5:

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

#### 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.  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.

#### Question 7:

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

#### 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.  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.

#### Question 9:

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

#### 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. 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.

#### Question 11:

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

Now, consider triangles BQX and APX.

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

Hence, proved.

#### 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. #### 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. ∴ AB = CQ            (CPCT)

Hence, proved.

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

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

#### 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.  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.

#### 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. Hence, proved.

#### Question 17:

In ABC, D is the midpoint of BC. If DLAB and DMAC such that DL = DM, prove that AB = AC. $\angle B=\angle C$
This makes triangle ABC an isosceles triangle.
Or AB = AC
Hence, proved.

#### 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. 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.

#### 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. 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

#### 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. 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)

#### 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. 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

#### 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. 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.

#### 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.  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)

#### Question 24:

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

#### 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. 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.

#### 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?

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 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?

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 1:

Is it possible to construct a triangle with lengths of its sides as given below? Give reason for your answer.
(i) 5 cm, 4 cm, 9 cm
(ii) 8 cm, 7 cm, 4 cm
(iii) 10 cm, 5 cm, 6 cm
(iv) 2.5 cm, 5 cm, 7 cm
(v) 3 cm, 4 cm, 8 cm

(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 2:

In ΔABC, ∠A = 50° and ∠B = 60°. Determine the longest and shortest sides of the triangle.

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.

#### Question 3:

(i) In ABC, A = 90°. Which is its longest side?
(ii)
In ABC, A = ∠B = 45°. Which is its longest side?
(iii) In ABC, A = 100° and ∠C = 50°. Which is its shortest side?

(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.

#### Question 4:

In ABC, side AB is produced to D such that BD = BC. If B = 60° and ∠A = 70°, prove that (i) AD > CD and (ii) AD > AC. In triangle CBA, CBD is an exterior angle.

Triangle BCD is isosceles and BC = BD.
Let .

#### Question 5:

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC.  Given: ∠B < ∠and ∠C < ∠D

Proof:

#### Question 6:

AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD. Show that ∠A > ∠C and
B > ∠D. 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:

#### Question 7:

In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).

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).

#### Question 8:

In a quadrilateral ABCD, show that
$\left(AB+BC+CD+DA\right)<2\left(BD+AC\right)$

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

In ∆BOC,

In ∆COD,

In ∆AOD,

#### Question 9:

In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.  Given: In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.

#### Question 10:

In the given figure, PQ > PR and QS and RS are the bisectors of Q and ∠R respectively. Show that SQ > SR. Since the angle opposite to the longer side is greater, we have:

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

$SQ>SR$

#### Question 11:

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD. Given: In $∆$ABC, AB = AC

To prove: CD BD

Proof:

In $∆$ABC,

Since, AB = AC       (Given)

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

#### Question 12:

Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle. 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.

#### Question 13:

In the given figure, prove that
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC.  To prove:
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC

Proof:
(i)

(ii)

#### Question 14:

If O is a point within ABC, show that:
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) $OA+OB+OC>\frac{1}{2}\left(AB+BC+CA\right)$ 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:

#### Question 15:

In the given figure, AD BC and CD > BD. Show that AC > AB.  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, $AC>AB$.

#### Question 16:

In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE. Prove that AB + AC > BE.  Given: CD DE

To prove: AB AC BE

Proof:

In $∆ABC$,

In $∆BED$,

From (1) and (2), we get

AB AC BE

#### Question 1:

Which of the following is not a criterion for congruence of triangles?
(a) SSA
(b) SAS
(c) ASA
(d) SSS

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

#### Question 2:

If AB = QR, BC = RP and CA = PQ, then which of the following holds?
(a) ABC ≅ ∆PQR
(b) CBA ≅ ∆PQR
(c) CAB ≅ ∆PQR
(d) BCA ≅ ∆PQR

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

$∆CAB\cong ∆PQR$

#### Question 3:

If ABC ≅ ∆PQR  then which of the following is not true?
(a) BC = PQ
(b) AC = PR
(c) BC = QR
(d) AB = PQ

(a) BC = PQ

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

#### Question 4:

In ABC, AB = AC and ∠B = 50°. Then, ∠A = ?
(a) 40°
(b) 50°
(c) 80°
(d) 130°

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).

#### Question 5:

In ABC, BC = AB and ∠B = 80°. Then, ∠A = ?
(a) 50°
(b) 40°
(c) 100°
(d) 80°

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).

#### Question 6:

In ABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?
(a) 4 cm
(b) 5 cm
(c) 8 cm
(d) 2.5 cm 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).

#### Question 7:

Two sides of a triangle are of length 4 cm and 2.5 cm. The length of the third side of the triangle cannot be
(a) 6 cm
(b) 6.5 cm
(c) 5.5 cm
(d) 6.3 cn

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).

#### Question 8:

In ABC, if ∠C > ∠B, then
(a) BC > AC
(b) AB > AC

(c) AB > AC
(d) BC > AC (b) AB > AC

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

#### Question 9:

It is given that ABC ≅ ∆ FDE in which AB = 5 cm, ∠B = 40°, A = 80° and FD = 5 cm. Then, which of the following is true?
(a) ∠D = 60°
(b) ∠E = 60°

(c) ∠F = 60°
(d) D = 80°

(b)​ $\angle E=60°$

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

$\angle E=60°$

#### Question 10:

In ABC, ∠A = 40° and ∠B = 60°. Then the longest side of ∆ABC is
(a) BC
(b) AC
(c) AB
(d) cannot be determined

(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.

#### Question 11:

In the given figure, AB > AC. Then which of the following is true?
(d) Cannot be determined $AB>AC$ is given.

$\angle ACB>\angle ABC$

#### Question 12:

In the given figure, AB > AC. If BO and CO are the bisectors of B and ∠C respectively, then
(a) OB = OC
(b) OB > OC
(c) OB < OC (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)
$⇒OB>OC$

#### Question 13:

In the given figure, AB = AC and OB = OC. Then, ABO : ∠ACO = ?
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) None of these (a) 1:1

​In

i.e., $\angle ABO=\angle ACO$
∴

#### Question 14:

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is
(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

(b) isosceles In

To prove: AB = AC

#### Question 15:

In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have
(a) A = ∠D
(b) ∠B = ∠E
(c) ∠C = ∠F
(d) none of these

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

#### Question 16:

In ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABC ≅ ∆DEF, we must have
(a) AB = DF
(b) AC = DE
(c) BC = EF
(d) ∠A = ∠D In order that $△ABC\cong △DEF$, we must have BC = EF.
Hence, the correct answer is option (c).

#### Question 17:

In ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles (a) isosceles but not congruent

Thus, both the triangles are isosceles but not congruent.

#### Question 18:

Which is true?
(a) A triangle can have two right angles.
(b) A triangle can have two obtuse angles.
(c) A triangle can have two acute angles.
(d) An exterior angle of a triangle is less than either of the interior opposite angles. (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.

#### Question 19:

Fill in the blanks with < or >.
(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) (Perimenter of a triangle) ...... (sum of its three medians)

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

#### Question 20:

Fill in the blanks.
(a) Each angle of an equilateral triangle measures ...... .
(b) Medians of an equilateral triangle are ...... .
(c) In a right triangle the hypotenuse is the ...... side.
(d) Drawing a ABC with AB = 3 cm, BC = 4 cm and CA = 7 cm is ...... .

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