Class 9 Exam > Class 9 Notes > Mathematics (Maths) Class 9 > RS Aggarwal Solutions: Congruence of Triangles and Inequalities in a Triangle- 1
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Page 1
Question:1
In the given figure, AB || CD and O is the midpoint of AD.
Show that
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Solution:
Given: In the given figure, AB || CD and O is the midpoint of AD.
To prove:
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Proof:
i
In ?AOB and ?DOC,
?BAO = ?CDO (Alternate interior angles, AB || CD)
AO = DO (Given, O is the midpoint of AD)
?AOB = ?DOC Verticallyoppositeangles
? By ASA congruence criteria,
?AOB ? ?DOC
ii
? ?AOB ? ?DOC
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.
Solution:
Given: In the given figure, AD and BC are equal perpendiculars to a line segment AB.
To prove: CD bisects AB
Proof:
In ?AOD and ?BOC,
?DAO = ?CBO = 90
°
Page 2
Question:1
In the given figure, AB || CD and O is the midpoint of AD.
Show that
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Solution:
Given: In the given figure, AB || CD and O is the midpoint of AD.
To prove:
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Proof:
i
In ?AOB and ?DOC,
?BAO = ?CDO (Alternate interior angles, AB || CD)
AO = DO (Given, O is the midpoint of AD)
?AOB = ?DOC Verticallyoppositeangles
? By ASA congruence criteria,
?AOB ? ?DOC
ii
? ?AOB ? ?DOC
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.
Solution:
Given: In the given figure, AD and BC are equal perpendiculars to a line segment AB.
To prove: CD bisects AB
Proof:
In ?AOD and ?BOC,
?DAO = ?CBO = 90
°
Given
AD = BC
Given
?DOA = ?COB
Verticallyoppositeangles
? By AAS congruence criteria,
?AOD ? ?BOC
? 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 ?ABC ? ?CDA.
Solution:
Given: In the given figure, two parallel line l and m are intersected by two parallel lines p and q.
To prove: ?ABC ? ?CDA
Proof:
In ?ABC and ?CDA,
?
BAC = ?
DCA (Alternate interior angles, p ?
q)
?
BCA = ?
DAC (Alternate interior angles, l ?
m)
AC = CA Commonside
?
By ASA congruence criteria,
?ABC ? ?CDA
Question:4
AD is an altitude of an isosceles ?ABC in which AB = AC.
Show that i
AD bisects BC, ii
AD bisects ?A.
Solution:
Given: AD is an altitude of an isosceles ?ABC in which AB = AC.
To prove: i
AD bisects BC, ii
AD bisects ?A
Page 3
Question:1
In the given figure, AB || CD and O is the midpoint of AD.
Show that
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Solution:
Given: In the given figure, AB || CD and O is the midpoint of AD.
To prove:
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Proof:
i
In ?AOB and ?DOC,
?BAO = ?CDO (Alternate interior angles, AB || CD)
AO = DO (Given, O is the midpoint of AD)
?AOB = ?DOC Verticallyoppositeangles
? By ASA congruence criteria,
?AOB ? ?DOC
ii
? ?AOB ? ?DOC
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.
Solution:
Given: In the given figure, AD and BC are equal perpendiculars to a line segment AB.
To prove: CD bisects AB
Proof:
In ?AOD and ?BOC,
?DAO = ?CBO = 90
°
Given
AD = BC
Given
?DOA = ?COB
Verticallyoppositeangles
? By AAS congruence criteria,
?AOD ? ?BOC
? 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 ?ABC ? ?CDA.
Solution:
Given: In the given figure, two parallel line l and m are intersected by two parallel lines p and q.
To prove: ?ABC ? ?CDA
Proof:
In ?ABC and ?CDA,
?
BAC = ?
DCA (Alternate interior angles, p ?
q)
?
BCA = ?
DAC (Alternate interior angles, l ?
m)
AC = CA Commonside
?
By ASA congruence criteria,
?ABC ? ?CDA
Question:4
AD is an altitude of an isosceles ?ABC in which AB = AC.
Show that i
AD bisects BC, ii
AD bisects ?A.
Solution:
Given: AD is an altitude of an isosceles ?ABC in which AB = AC.
To prove: i
AD bisects BC, ii
AD bisects ?A
Proof:
i
In ?ABD and ?ACD,
?
ADB = ?
ADC = 90°
(Given, AD ?
BC)
AB = AC Given
AD = AD Commonside
?
By RHS congruence criteria,
?ABD ? ?ACD
?
BD = CD CPCT
Hence, AD bisects BC.
ii
?
?ABD ? ?ACD
From(i)
?
?
BAD = ?
CAD CPCT
Hence, AD bisects ?A.
Question:5
In the given figure, BE and CF are two equal altitudes of ?ABC.
Show that i
?ABE ? ?ACF, ii
AB = AC.
Solution:
In ? ABE and ? ACF, we have: BE = CF (Given) ?BEA = ?CFA = 90° ?A = ?A (Common) ? ABE ? ? ACF (AAS criterion)
AB = AC CPCT
Question:6
?ABC and ?DBC 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
?ABD ? ?ACD
ii
?ABE ? ?ACE
iii
AE bisects ?A as well as ?D
iv
AE is the perpendicular bisector of BC.
Solution:
Given: ?ABC and ?DBC are two isosceles triangles on the same base BC.
To prove:
i
?ABD ? ?ACD
ii
?ABE ? ?ACE
Page 4
Question:1
In the given figure, AB || CD and O is the midpoint of AD.
Show that
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Solution:
Given: In the given figure, AB || CD and O is the midpoint of AD.
To prove:
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Proof:
i
In ?AOB and ?DOC,
?BAO = ?CDO (Alternate interior angles, AB || CD)
AO = DO (Given, O is the midpoint of AD)
?AOB = ?DOC Verticallyoppositeangles
? By ASA congruence criteria,
?AOB ? ?DOC
ii
? ?AOB ? ?DOC
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.
Solution:
Given: In the given figure, AD and BC are equal perpendiculars to a line segment AB.
To prove: CD bisects AB
Proof:
In ?AOD and ?BOC,
?DAO = ?CBO = 90
°
Given
AD = BC
Given
?DOA = ?COB
Verticallyoppositeangles
? By AAS congruence criteria,
?AOD ? ?BOC
? 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 ?ABC ? ?CDA.
Solution:
Given: In the given figure, two parallel line l and m are intersected by two parallel lines p and q.
To prove: ?ABC ? ?CDA
Proof:
In ?ABC and ?CDA,
?
BAC = ?
DCA (Alternate interior angles, p ?
q)
?
BCA = ?
DAC (Alternate interior angles, l ?
m)
AC = CA Commonside
?
By ASA congruence criteria,
?ABC ? ?CDA
Question:4
AD is an altitude of an isosceles ?ABC in which AB = AC.
Show that i
AD bisects BC, ii
AD bisects ?A.
Solution:
Given: AD is an altitude of an isosceles ?ABC in which AB = AC.
To prove: i
AD bisects BC, ii
AD bisects ?A
Proof:
i
In ?ABD and ?ACD,
?
ADB = ?
ADC = 90°
(Given, AD ?
BC)
AB = AC Given
AD = AD Commonside
?
By RHS congruence criteria,
?ABD ? ?ACD
?
BD = CD CPCT
Hence, AD bisects BC.
ii
?
?ABD ? ?ACD
From(i)
?
?
BAD = ?
CAD CPCT
Hence, AD bisects ?A.
Question:5
In the given figure, BE and CF are two equal altitudes of ?ABC.
Show that i
?ABE ? ?ACF, ii
AB = AC.
Solution:
In ? ABE and ? ACF, we have: BE = CF (Given) ?BEA = ?CFA = 90° ?A = ?A (Common) ? ABE ? ? ACF (AAS criterion)
AB = AC CPCT
Question:6
?ABC and ?DBC 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
?ABD ? ?ACD
ii
?ABE ? ?ACE
iii
AE bisects ?A as well as ?D
iv
AE is the perpendicular bisector of BC.
Solution:
Given: ?ABC and ?DBC are two isosceles triangles on the same base BC.
To prove:
i
?ABD ? ?ACD
ii
?ABE ? ?ACE
iii
AE bisects ?A as well as ?D
iv
AE is the perpendicular bisector of BC
Proof:
i
In ?ABD and ?ACD,
BD = CD (Given, ?DBC is an isosceles triangles)
AB = AC (Given, ?ABC is an isosceles triangles)
AD = AD Commonside
?
By SSS congruence criteria,
?ABD ? ?ACD
Also, ?
BAD = ?
CAD CPCT
or, ?
BAE = ?
CAE .....1
ii
In ?ABE and ?ACE,
AB = AC (Given, ?ABC is an isosceles triangles)
?
BAE = ?
CAE
From(i)
AE = AE Commonside
?
By SAS congruence criteria,
?ABE ? ?ACE
Also, BE = CE CPCT
.....2
And, ?
AEB = ?
AEC CPCT
.....3
iii
In ?BED and ?CED,
BD = CD (Given, ?DBC is an isosceles triangles)
BE = CE
From(2)
DE = DE Commonside
?
By SSS congruence criteria,
?BED ? ?CED
Also, ?
BDE = ?
CDE CPCT
.....4
iv
?
?
BAE = ?
CAE
From(1)
And, ?
BDE = ?
CDE
From(4)
?
AE bisects ?A as well as ?D.
v
Page 5
Question:1
In the given figure, AB || CD and O is the midpoint of AD.
Show that
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Solution:
Given: In the given figure, AB || CD and O is the midpoint of AD.
To prove:
i
?AOB ? ?DOC.
ii
O is the midpoint of BC.
Proof:
i
In ?AOB and ?DOC,
?BAO = ?CDO (Alternate interior angles, AB || CD)
AO = DO (Given, O is the midpoint of AD)
?AOB = ?DOC Verticallyoppositeangles
? By ASA congruence criteria,
?AOB ? ?DOC
ii
? ?AOB ? ?DOC
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.
Solution:
Given: In the given figure, AD and BC are equal perpendiculars to a line segment AB.
To prove: CD bisects AB
Proof:
In ?AOD and ?BOC,
?DAO = ?CBO = 90
°
Given
AD = BC
Given
?DOA = ?COB
Verticallyoppositeangles
? By AAS congruence criteria,
?AOD ? ?BOC
? 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 ?ABC ? ?CDA.
Solution:
Given: In the given figure, two parallel line l and m are intersected by two parallel lines p and q.
To prove: ?ABC ? ?CDA
Proof:
In ?ABC and ?CDA,
?
BAC = ?
DCA (Alternate interior angles, p ?
q)
?
BCA = ?
DAC (Alternate interior angles, l ?
m)
AC = CA Commonside
?
By ASA congruence criteria,
?ABC ? ?CDA
Question:4
AD is an altitude of an isosceles ?ABC in which AB = AC.
Show that i
AD bisects BC, ii
AD bisects ?A.
Solution:
Given: AD is an altitude of an isosceles ?ABC in which AB = AC.
To prove: i
AD bisects BC, ii
AD bisects ?A
Proof:
i
In ?ABD and ?ACD,
?
ADB = ?
ADC = 90°
(Given, AD ?
BC)
AB = AC Given
AD = AD Commonside
?
By RHS congruence criteria,
?ABD ? ?ACD
?
BD = CD CPCT
Hence, AD bisects BC.
ii
?
?ABD ? ?ACD
From(i)
?
?
BAD = ?
CAD CPCT
Hence, AD bisects ?A.
Question:5
In the given figure, BE and CF are two equal altitudes of ?ABC.
Show that i
?ABE ? ?ACF, ii
AB = AC.
Solution:
In ? ABE and ? ACF, we have: BE = CF (Given) ?BEA = ?CFA = 90° ?A = ?A (Common) ? ABE ? ? ACF (AAS criterion)
AB = AC CPCT
Question:6
?ABC and ?DBC 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
?ABD ? ?ACD
ii
?ABE ? ?ACE
iii
AE bisects ?A as well as ?D
iv
AE is the perpendicular bisector of BC.
Solution:
Given: ?ABC and ?DBC are two isosceles triangles on the same base BC.
To prove:
i
?ABD ? ?ACD
ii
?ABE ? ?ACE
iii
AE bisects ?A as well as ?D
iv
AE is the perpendicular bisector of BC
Proof:
i
In ?ABD and ?ACD,
BD = CD (Given, ?DBC is an isosceles triangles)
AB = AC (Given, ?ABC is an isosceles triangles)
AD = AD Commonside
?
By SSS congruence criteria,
?ABD ? ?ACD
Also, ?
BAD = ?
CAD CPCT
or, ?
BAE = ?
CAE .....1
ii
In ?ABE and ?ACE,
AB = AC (Given, ?ABC is an isosceles triangles)
?
BAE = ?
CAE
From(i)
AE = AE Commonside
?
By SAS congruence criteria,
?ABE ? ?ACE
Also, BE = CE CPCT
.....2
And, ?
AEB = ?
AEC CPCT
.....3
iii
In ?BED and ?CED,
BD = CD (Given, ?DBC is an isosceles triangles)
BE = CE
From(2)
DE = DE Commonside
?
By SSS congruence criteria,
?BED ? ?CED
Also, ?
BDE = ?
CDE CPCT
.....4
iv
?
?
BAE = ?
CAE
From(1)
And, ?
BDE = ?
CDE
From(4)
?
AE bisects ?A as well as ?D.
v
? ?AEB + ?AEC = 180° (Linear pair) ? ?AEB + ?AEB = 180° [From (3)] ? 2 ?AEB = 180° ? ?AEB =
180°
2
? ?AEB = 90° . . . . . (5)
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.
Solution:
Consider the triangles AEB and CDB.
?EBA = ?DBC
Commonangle
...i
Further, we have:
?BEA = 180 -y ?BDC = 180 -xSince x = y, we have: 180 -x = 180 -y ? ?BEA = ?BDC . . . (ii)
AB = CB Given
...iii
From i
, ii
and iii
, we have:
? BDC ? ? BEA
AAScriterion
? 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
?APB ? ?AQB
ii
BP = BQ, i.e., B is equidistant from the arms of ?A.
Solution:
Given: In the given figure, ?BAQ = ?BAP, BP ?
AP and BQ ?
AQ.
To prove:
i
?APB ? ?AQB
ii
BP = BQ, i.e., B is equidistant from the arms of ?A.
Proof:
i
In ?APB and ?AQB,
?BAQ = ?BAP, Given
?APB = ?AQB = 90°
(Given, BP ?
AP and BQ ?
AQ)
AB = AB Commonside
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FAQs on RS Aggarwal Solutions: Congruence of Triangles and Inequalities in a Triangle- 1 - Mathematics (Maths) Class 9
| 1. What are congruent triangles? |  |
Ans. Congruent triangles are two triangles that have the exact same shape and size. This means that all corresponding sides and angles of the triangles are equal.
| 2. How can we prove that two triangles are congruent? | |
Ans. There are several ways to prove that two triangles are congruent. Some common methods include using the side-side-side (SSS) congruence criterion, the side-angle-side (SAS) congruence criterion, and the angle-side-angle (ASA) congruence criterion.
| 3. What are the inequalities in a triangle? | |
Ans. The inequalities in a triangle refer to the relationships between the lengths of the sides and the measures of the angles in a triangle. For example, the length of any side of a triangle must be less than the sum of the lengths of the other two sides.
| 4. How can we prove inequalities in a triangle? | |
Ans. One way to prove inequalities in a triangle is by using the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Another way is by using the properties of angles in a triangle, such as the fact that the measure of any angle in a triangle must be less than 180 degrees.
| 5. How are congruence of triangles and inequalities in a triangle related? | |
Ans. Congruence of triangles and inequalities in a triangle are related because the congruence criteria often involve comparing the lengths of sides and the measures of angles in triangles. Inequalities in triangles help establish the conditions under which triangles can be proven congruent. Additionally, inequalities in triangles can be used to determine whether a given set of side lengths and angle measures can form a valid triangle.
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Nov 22, 2024 Last updated |
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