These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1
Question 1.
In quadrilateral ACBD (See Fig. 7.16), AC = AD, and AB bisects ∠A. Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
Solution:
In ΔABC and ΔABD
AC = AD (Given)
∠CAB = ∠DAB (Given)
AB = AB (Common)
Therefore, By SAS congruency condition
ΔABC ≅ ΔABD
So, BC = BD (By C.P.C.T)
Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17)
Prove that:
(i) ΔABD = ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC
Solution:
(i) In ΔABD and ΔBAC
AD = BC
∠DAB = ∠CBA (Given)
and AB = BA (Common)
By SAS Congruency Condition
ΔABD = ΔBAC
(ii) BD= AC (By C.P.C.T)
(iii) ∠ABD = ∠BAC (Again by C.P.C.T)
Question 3.
AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
Solution:
In ΔAOD and ΔBOC,
AD = BC (Given)
∠OAD = ∠OBC (each 90°)
∠AOD = ∠BOC (Vertically opposite angles)
Therefore, by ASA congruency condition.
ΔAOD = ΔBOC
So, OA = OB (by C.P.C.T)
Hence, CD bisects line segment AB.
Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC = ΔCDA.
Solution:
We have given that l || m and p || q
Therefore, In ΔABC and ΔCDA
∠BAC = ∠DCA
(Alternate interior angles as AB || DC)
∠ACB = ∠CAD
(Alternate interior angles as BC || DA)
AC = CA (common)
So, By A-S-A congruency condition,
ΔABC = ΔCDA
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 Fig. 7.20) show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistance from the arms of ∠A.
Solution:
In ΔABP and ΔABQ,
∠BAP = ∠BAQ (Given)
∠APB = ∠AQB (Each 90°)
AB = AB (Common)
By A-A-S congruency condition.
So, ΔABP ≅ ΔABQ
(ii) BP = BQ (By C.P.C.T)
Question 6.
In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution:
In ΔBAC and ΔDAE
AB = AD (Given)
AC = AE (Given)
∠BAD = ∠EAC ……(i) (Given)
Adding ∠DAC both side in equation (i)
∴ ∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
Therefore by S-A-S Congruency Condition
ΔBAC = ΔDAE
So, BO = DE (By C.P.C.T)
Question 7.
AB is a line segment and P is its midpoint. 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) ΔDAP ≅ ΔEBP
(ii) AD = BE
Solution:
(i) In ΔDAP and ΔEBP
∠DAP= ∠EBP (Given)
∠APE = ∠DPB (Given)
∴ ∠APE + ∠EPD = ∠DPB + ∠EPD (Add ∠EPD both side)
∠APD = ∠BPE
AP = BP (Given P is the mid point of AB)
∴ By A-S-A Congruency Condition,
ΔDAP ≅ ΔEBP
(ii) AD = BE (By C.P.C.T.)
Question 8.
In the right triangle ABC right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produces to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = \(\frac {1}{2}\) AB
Solution:
(i) In ΔAMC and ΔBMD,
AM = BM (Given)
CM = DM (Given)
∠AMC = ∠BMD (Vertically opposite angles)
∴ By S-A-S Congruency Condition.
ΔAMC = ΔBMD
(ii) ∠CAM = ∠DBM (by C.P.C.T)
Also, ∠CAM + ∠MBC = 90° (Since ∠C = 90°)
∴ ∠DBM + ∠MBC = 90° (∵ ∠CAM = ∠DBM)
or, ∠DBC = 90°
(iii) In ΔDBC and ΔACB,
BC = BC (Common)
DB = AC (∵ ΔBMD ≅ ΔAMC by C.PC.T)
and ∠DBC = ∠ACB (each 90° proved above)
Therefore, by S-A-S Congruency Condition,
∴ ΔDBC ≅ ΔACB
(iv) Since, ΔDBC ≅ ΔACB,
DC = AB
∴ \(\frac {1}{2}\) DC = \(\frac {1}{2}\) AB
CM = AM
or CM = \(\frac {1}{2}\) AB