![myCBSEguide App](https://media-mycbseguide.s3.amazonaws.com/images/direct/book_round_corner.png)
myCBSEguide App
Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.
Install NowNCERT Solutions for Class 9 Maths Exercise 7.1 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.
NCERT solutions for Class 9 Maths Triangles Download as PDF
NCERT Solutions for Class 9 Mathematics Triangles
1. In quadrilateral ABCD (See figure). AC = AD and AB bisects
A. Show that
ABC
ABD. What can you say about BC and BD?
Ans. Given: In quadrilateral ABCD, AC = AD and AB bisects A.
To prove: ABC
ABD
Proof: In ABC and
ABD,
AC = AD [Given]
BAC =
BAD [
AB bisects
A]
AB = AB [Common]
ABC
ABD [By SAS congruency]
Thus BC = BD [By C.P.C.T.]
2. ABCD is a quadrilateral in which AD = BC and
DAB =
CBA. (See figure). Prove that:
(i) ABD
BAC
(ii) BD = AC
(iii) ABD =
BAC
Ans. (i) In ABC and
ABD,
BC = AD [Given]
DAB =
CBA [Given]
AB = AB [Common]
ABC
ABD [By SAS congruency]
Thus AC = BD [By C.P.C.T.]
(ii) Since ABC
ABD
AC = BD [By C.P.C.T.]
(iii) Since ABC
ABD
ABD =
BAC [By C.P.C.T.]
3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB (See figure)
Ans. In BOC and
AOD,
OBC =
OAD =
[Given]
BOC =
AOD [Vertically Opposite angles]
BC = AD [Given]
BOC
AOD [By ASA congruency]
OB = OA and OC = OD [By C.P.C.T.]
NCERT Solutions for Class 9 Maths Exercise 7.1
4.
and
are two parallel lines intersected by another pair of parallel lines
and
(See figure). Show that
ABC
CDA.
Ans. AC being a transversal. [Given]
Therefore DAC =
ACB [Alternate angles]
Now [Given]
And AC being a transversal. [Given]
Therefore BAC =
ACD [Alternate angles]
Now In ABC and
ADC,
ACB =
DAC [Proved above]
BAC =
ACD [Proved above]
AC = AC [Common]
ABC
CDA [By ASA congruency]
NCERT Solutions for Class 9 Maths Exercise 7.1
5. Line
is the bisector of the angle A and B is any point on
BP and BQ are perpendiculars from B to the arms of
A. Show that:
(i) APB
AQB
(ii) BP = BQ or P is equidistant from the arms of A (See figure).
Ans. Given: Line bisects
A.
BAP =
BAQ
(i) In ABP and
ABQ,
BAP =
BAQ [Given]
BPA =
BQA =
[Given]
AB = AB [Common]
APB
AQB [By ASA congruency]
(ii) Since APB
AQB
BP = BQ [By C.P.C.T.]
B is equidistant from the arms of
A.
NCERT Solutions for Class 9 Maths Exercise 7.1
6. In figure, AC = AB, AB = AD and
BAD =
EAC. Show that BC = DE.^
Ans. Given that BAD =
EAC
Adding DAC on both sides, we get
BAD +
DAC =
EAC +
DAC
BAC =
EAD ……….(i)
Now in ABC and
AED,
AB = AD [Given]
AC = AE [Given]
BAC =
DAE [From eq. (i)]
ABC
ADE [By SAS congruency]
BC = DE [By C.P.C.T.]
NCERT Solutions for Class 9 Maths Exercise 7.1
7. AB is a line segment and P is the mid-point. D and E are points on the same side of AB such that
BAD =
ABE and
EPA =
DPB. Show that:
(i) DAF
FBP
(ii) AD = BE (See figure)
Ans. Given that EPA =
DPB
Adding EPD on both sides, we get
EPA +
EPD =
DPB +
EPD
APD =
BPE ……….(i)
Now in APD and
BPE,
PAD =
PBE [
BAD =
ABE (given),
PAD =
PBE]
AP = PB [P is the mid-point of AB]
APD =
BPE [From eq. (i)]
DPA
EBP [By ASA congruency]
AD = BE [ By C.P.C.T.]
NCERT Solutions for Class 9 Maths Exercise 7.1
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 figure)
Show that:
(i) AMC
BMD
(ii) DBC is a right angle.
(iii) DBC
ACB
(iv) CM = AB
Ans. (i) In AMC and
BMD,
AM = BM [AB is the mid-point of AB]
AMC =
BMD [Vertically opposite angles]
CM = DM [Given]
AMC
BMD [By SAS congruency]
ACM =
BDM ……….(i)
CAM =
DBM and AC = BD [By C.P.C.T.]
(ii) For two lines AC and DB and transversal DC, we have,
ACD =
BDC [Alternate angles]
AC
DB
Now for parallel lines AC and DB and for transversal BC.
DBC =
ACB [Alternate angles] ……….(ii)
But ABC is a right angled triangle, right angled at C.
ACB =
……….(iii)
Therefore DBC =
[Using eq. (ii) and (iii)]
DBC is a right angle.
(iii) Now in DBC and
ABC,
DB = AC [Proved in part (i)]
DBC =
ACB =
[Proved in part (ii)]
BC = BC [Common]
DBC
ACB [By SAS congruency]
(iv) Since DBC
ACB [Proved above]
DC = AB
AM + CM = AB
CM + CM = AB [
DM = CM]
2CM = AB
CM =
AB
NCERT Solutions for Class 9 Maths Exercise 7.1
NCERT Solutions for Class 9 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 9 Maths have total 15 chapters. 9 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 9 solutions PDF and Maths ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.
CBSE app for Class 9
To download NCERT Solutions for Class 9 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through the best app for CBSE students and myCBSEguide website.
![myCBSEguide App](https://media-mycbseguide.s3.amazonaws.com/images/Logo.png)
Test Generator
Create question paper PDF and online tests with your own name & logo in minutes.
Create Now![myCBSEguide App](https://media-mycbseguide.s3.amazonaws.com/images/direct/book_round_corner.png)
myCBSEguide
Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes
Install Now
This is very easy language.
It’s is easy and simple for understanding. Thankyou very much ?
This is very easy for understanding
Thanks sir very helpful
Nice website
Thank. you
Very easy language used
Nice mathmatician, you should have in the top of the tight corner
Excellent ! Fantastic