### 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

To proveABC ABD
Proof: In ABC and ABD,
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,

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,

BOC = AOD [Vertically Opposite angles]

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]

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.

Adding DAC on both sides, we get

BAD + DAC = EAC + DAC

AC = AE [Given]

BAC = DAE [From eq. (i)]

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,

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.
$\mathrm{\angle }DBC+\mathrm{\angle }ACB={180}^{0}$[cointerior angles].....(ii)
But ABC is a right angled triangle, right angled at C.
ACB = ${90}^{0}$ ……….(iii)
Therefore DBC = ${90}^{0}$ [Using eq. (ii) and (iii)]
DBC is a right angle.

(iii) Now in DBC and ABC,

DB = AC [Proved in part (i)]
DBC = ACB = ${90}^{0}$ [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 = $\frac{1}{2}$ AB