NCERT Class 9 Maths Solutions Chapter - 7 Triangles, Ex - 7.1

Ex - 7.1

Question 1.  In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC  ABD. What can you say about BC and BD?


Solution

In ABC and ABD
AC = AD                                            (given)
CAB = DAB                                 (given)
AB = AB                                             (common)

So, BC and BD are of equal length.

Question 2.  ABCD is a quadrilateral in which AD = BC and DAB = CBA (See the given figure). Prove that

(i)    ABD  BAC

(ii)    BD = AC                

(iii)    ABD = BAC

Solution

In ABD and BAC
    AD = BC                                         (given)
   DAB = CBA                              (given)
    AB = BA                                         (common)

And ABD = BAC                          (by CPCT)

Question 3.  AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.

Solution

In BOC and AOD
BOC = AOD                                 (vertically opposite angles)
CBO = DAO                                 (each 90^o)
BC = AD                                             (given)

Question 4.  l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ABC  CDA.

Solution

Question 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 the given figure). Show that:

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

Solution

Question 6.  In the given figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.

Solution

Given that BAD = EAC 
BAD + DAC = EAC + DAC
BAC = DAE 
Now in BAC and DAE
AB = AD                                             (given)
BAC = DAE                                 (proved above)
AC = AE                                             (given)

Question 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 the given figure). Show that 

(i) DAP  EBP
(ii) AD = BE

Solution

Given that EPA = DPB

 EPA + DPE = DPB + DPE
 DPA = EPB
     Now in  DAP and  EBP
    DAP = EBP                               (given)
    AP = BP                                          (P is mid point of AB)
    DPA = EPB                              (from above)

Question 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 the given figure).  Show that:

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

Solution

 (i)  In AMC and BMD
AM = BM                                              (M is mid point of AB)
AMC = BMD                                  (vertically opposite angles)
CM = DM                                             (given)

(ii) We have ACM = BDM
But ACM and BDM are alternate interior angles
Since alternate angles are equal.
Hence, we can say that DB || AC
 DBC + ACB = 180^o                   (co-interior angles) DBC + 90^o = 180^o
 DBC + 90^o = 180⁰ 

DBC          =  90^o

(iii) Now in DBC and ACB
DB = AC                                             (Already proved)
DBC = ACB                                 (each 90^o )
BC = CB                                             (Common)

(iv) We have DBC  ACB

 

Ex-6.3

Ex-7.2