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