Triangles - Solutions
CBSE Class 9 Mathematics
NCERT Solutions
CHAPTER 7
Triangles(Ex. 7.1)
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 BAD,
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 AAS congruency]
OB = OA [By C.P.C.T., Also, OC = OD again by C.P.C.T.]
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]
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 B 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 AAS congruency]
(ii) Since APB AQB
BP = BQ [By C.P.C.T.]
B is equidistant from the arms of A.
6. In figure, AC = AE, 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 ADE,
AB = AD [Given]
AC = AE [Given]
BAC = DAE [From eq. (i)]
ABC ADE [By SAS congruency]
BC = DE [By C.P.C.T.]
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) DAP EBP
(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)]
DAP EBP [By ASA congruency]
AD = BE [ By C.P.C.T.]
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 [M 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.
[cointerior 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 DM + CM = AB CM + CM = AB [ DM = CM] 2CM = AB CM = AB
