Exercise 7.1 maths class 9 || maths chapter 7 class 9 exercise 7.1 || exercise 7.1 class 9 maths || rd sharma class 9 triangle || chapter 7 maths class 9 exercise 7.1 || class 9 ncert exercise 7.1 || chapter 7 exercise 7.1 class 9 || class 9 maths chapter 7 exercise 7.1 solutions || ncert class 9 exercise 7.1
Explore detailed, step-by-step solutions for Class 9 Maths Chapter 7, Exercise 7.1, centered on the topic of Triangles. This exercise introduces students to the fundamental properties and classification of triangles based on sides and angles. It lays the groundwork for understanding congruence in triangles by guiding learners through problems involving angle sum property, exterior angle theorem, and basic triangle properties. By working through Exercise 7.1, students strengthen their logical thinking, apply geometric rules systematically, and build a solid conceptual base for more advanced topics such as congruence and inequalities in triangles in later chapters and higher classes.

exercise 7.1 maths class 9 || maths chapter 7 class 9 exercise 7.1 || exercise 7.1 class 9 maths || rd sharma class 9 triangle || chapter 7 maths class 9 exercise 7.1 || class 9 ncert exercise 7.1 || chapter 7 exercise 7.1 class 9 || class 9 maths chapter 7 exercise 7.1 solutions || ncert class 9 exercise 7.1
Exercise 7.1
1. In quadrilateral ACBD,AC=AD
and AB bisect A (see Fig. 7.16). Show that △ABC △ABD. What can you say about BC and BD?


Answer
It is given that AC and AD are equal
i.e. AC=AD and the line segment AB bisects A.We will have to now prove that the two triangles ABC and ABD are similar i.e. △ABCΔABD
Proof:
Consider the triangles △ABC and △ABD,
(i) AC=AD (It is given in the question)
(ii) AB=AB( Common)
(iii) CAB=DAB (Since AB is the bisector of angle A )
So, by SAS congruency criterion, △ABC△ABD.
For the 2nd part of the question, BC and BD are of equal lengths by the rule of C.P.C.T.
2. ABCD is a quadrilateral in which AD=BC and DAB=CBA
(see Fig. 7.17). Prove that
(i) △ABD△BAC (ii) BD=AC (iii) ∠ABD=∠BAC.

(i) △ABD△BAC (ii) BD=AC (iii) ∠ABD=∠BAC.

Answer
The given parameters from the questions are DAB=CBA and AD=BC.(i) △ABD and △BAC are similar by SAS congruency as
AB=BA (It is the common arm)
DAB=CBA and AD=BC (These are given in the question)
So, triangles ABD and BAC are similar i.e. △ABD△BAC. (Hence proved).
(ii) It is now known that △ABD△BAC so, BD=AC (by the rule of CPCT).
(iii) Since △ABD△BAC so,
Angles ABD=BAC (by the rule of CPCT ).
3. AD and BC are equal perpendiculars to a line
segment AB (see Fig. 7.18). Show that CD bisects
AB.


Answer
It is given that AD and BC are two equal
perpendiculars to AB.We will have to prove that CD is the bisector of AB
Now,
Triangles △AOD and △BOC are similar by AAS congruency since:
(i) A=B (They are perpendiculars)
(ii) AD=BC (As given in the question)
(iii) AOD=BOC (They are vertically opposite angles)
∴△AOD△BOC.
So, AO=OB (by the rule of CPCT ).
Thus, CD bisects AB (Hence proved).
exercise 7.1 maths class 9 || maths chapter 7 class 9 exercise 7.1 || exercise 7.1 class 9 maths || rd sharma class 9 triangle || chapter 7 maths class 9 exercise 7.1 || class 9 ncert exercise 7.1 || chapter 7 exercise 7.1 class 9 || class 9 maths chapter 7 exercise 7.1 solutions || ncert class 9 exercise 7.1
4. l and m are two parallel lines intersected by
another pair of parallel lines p and q (see Fig.
7.19). Show that △ABC△CDA.


Answer
It is given that p q and 1 mTo prove:
Triangles ABC and CDA are similar i.e. △ABC△CDA
Proof:
Consider the △ABC and △CDA,
(i) BCA=DAC and BAC=DCA Since they are alternate interior angles
(ii) AC=CA as it is the common arm
So, by ASA congruency criterion, △ABC△CDA.
5. Line l is the bisector of an angle A and B
is any point on l. BP and BQ are perpendiculars
from B to the arms of A (see Fig. 7.20). Show that:
(i) △APB△AQB
(ii) BP=BQ or B is equidistant from the arms of A .

(i) △APB△AQB
(ii) BP=BQ or B is equidistant from the arms of A .

Answer
It is given that the line "l" is the bisector of angle A and
the line segments BP and BQ are perpendiculars drawn from l.(i) △APB and △AQB are similar by AAS congruency because:
P=Q (They are the two right angles)
AB=AB (It is the common arm)
BAP=BAQ (As line l is the bisector of angle A)
So, △APB△AQB.
(ii) By the rule of CPCT,BP=BQ. So, it can be said the point B is equidistant from the arms of A .
6. In Fig. 7.21, AC=AE,AB=AD and BAD=EAC.
Show that BC=DE.


Answer
It is given in the question that AB=AD,AC=AE, and ∠BAD=∠EACTo prove:
The line segment BC and DE are similar i.e. BC=DE
Proof:
We know that BAD=EAC
Now, by adding DAC on both sides we get,
BAD+DAC=EAC+DAC
This implies, BAC=EAD
Now, △ABC and △ADE are similar by SAS congruency since:
(i) AC=AE (As given in the question)
(ii) BAC=EAD
(iii) AB=AD (It is also given in the question)
∴ Triangles ABC and ADE are similar i.e. △ABC△ADE.
So, by the rule of CPCT, it can be said that BC=DE.
7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that
BAD=ABE and EPA=DPB (see Fig. 7.22). Show
that
(i) △DAP △EBP
(ii) AD=BE

(i) △DAP △EBP
(ii) AD=BE

Answer
In the question, it is given that P is the mid-point of line
segment AB. Also, BAD=ABE and EPA=DPB(i) It is given that EPA=DPB
Now, add DPE on both sides,
EPA+DPE=DPB+DPE
This implies that angles DPA and EPB are equal i.e. DPA=EPB
Now, consider the triangles DAP and EBP.
DPA=EPB
AP=BP (Since P is the mid-point of the line segment AB)
BAD=ABE (As given in the question)
So, by ASA congruency, △DAP△EBP.
(ii) By the rule of CPCT,AD=BE.
8. In right triangle ABC, right angled at C,M is the
mid-point of hypotenuse AB. C is joined to M
and produced to a point D such that DM=CM. Point D
is joined to point B (see Fig. 7.23). Show that:
(i) △AMC△BMD
(ii) DBC is a right angle.
(iii) △DBC△ACB
(iv) CM=12AB

(i) △AMC△BMD
(ii) DBC is a right angle.
(iii) △DBC△ACB
(iv) CM=12AB

Answer
It is given that M is the mid-point of the line segment AB,C=90∘, and DM=CM(i) Consider the triangles △AMC and △BMD :
AM=BM (Since M is the mid-point)
CM=DM (Given in the question)
CMA=DMB (They are vertically opposite angles)
So, by SAS congruency criterion, △AMC△BMD.
(ii) ACM=BDM (by CPCT)
∴ACBD as alternate interior angles are equal.
Now, ACB+DBC=180∘ (Since they are co-interiors angles)
⇒90∘+B=180∘
∴DBC=90∘
(iii) In △DBC and △ACB,
BC=CB (Common side)
ACB=DBC (They are right angles)
DB=AC (by CPCT)
So, △DBC△ACB by SAS congruency.
(iv) DC=AB (Since △DBC△ACB )
⇒DM=CM=AM=BM (Since M the is mid-point)
So, DM+CM=BM+AM
Hence, CM+CM=AB
⇒CM=(12)AB