# CLASS 9 MATHS CHAPTER-7 TRIANGLES

### Exercise 7.1

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

Solution:

To prove: ABC 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.]

Question 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

Solution:
(i)In ABC and ABD,

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.]

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

Solution:
In BOC and AOD,

BOC = AOD [Vertically Opposite angles]

BOC AOD [By ASA congruency]

OB = OA and OC = OD [By C.P.C.T.]

Question 4.
and are two parallel lines intersected by another pair of parallel lines and (See figure). Show that ABC CDA.

Solution:
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]

Question 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 P is equidistant from the arms of A (See figure).

Solution:
Given: Line bisects A.

BAP = BAQ

(i) In ABP and ABQ,

BAP = BAQ [Given]

BPA = BQA = [Given]

AB = AB [Common]

APB AQB [By ASA congruency]

(ii) Since APB AQB

BP = BQ [By C.P.C.T.]

B is equidistant from the arms of A.

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

Solution:

Adding DAC on both sides, we get

BAD + DAC = EAC + DAC

Now in ABC and AED,

AC = AE [Given]

BAC = DAE [From eq. (i)]

BC = DE [By C.P.C.T.]

Question 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) DAF FBP

(ii) AD = BE (See figure)

Solution:

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)]

DPA EBP [By ASA congruency]

AD = BE [ By C.P.C.T.]

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 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 [AB 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.

DBC = ACB [Alternate 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

AM + CM = AB

CM + CM = AB [DM = CM]

2CM = AB

CM = AB

### NCERT Solutions for Class 9 Maths Exercise 7.2

Question 1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that
(i) OB = OC
(ii) AO bisects ∠A

Solution:
i) in ∆ABC, we have
AB = AC [Given]
∴∠ABC = ∠ACB [Angles opposite to equal sides of a A are equal]

∠ABC = ∠ACB
or ∠OBC = ∠OCB
⇒OC = OB [Sides opposite to equal angles of a ∆ are equal]

(ii) In ∆ABO and ∆ACO, we have
AB = AC [Given]
∠OBA = ∠OCA [ ∵∠B = ∠C]
OB = OC [Proved above]
∆ABO ≅∆ACO [By SAS congruency]
⇒∠OAB = ∠OAC [By C.P.C.T.]
⇒AO bisects ∠A.

Question 2. In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.

Solution:
Since AD is bisector of BC.
∴BD = CD
Now, in ∆ABD and ∆ACD, we have
BD = CD [Proved above]
∴∆ABD ≅∆ACD [By SAS congruency]
⇒AB = AC [By C.P.C.T.]
Thus, ∆ABC is an isosceles triangle.

Question 3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

Solution:
∆ABC is an isosceles triangle.
∴AB = AC
⇒∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal]
⇒∠BCE = ∠CBF
Now, in ∆BEC and ∆CFB
∠BCE = ∠CBF [Proved above]
∠BEC = ∠CFB [Each 90°]
BC = CB [Common]
∴∆BEC ≅∆CFB [By AAS congruency]
So, BE = CF [By C.P.C.T.]

Question 4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure).

Show that
(i) ∆ABE ≅∆ACF
(ii) AB = AC i.e., ABC is an isosceles triangle.

Solution:
(i) In ∆ABE and ∆ACE, we have
∠AEB = ∠AFC
[Each 90° as BE ⊥AC and CF ⊥AB]
∠A = ∠A [Common]
BE = CF [Given]
∴∆ABE ≅∆ACF [By AAS congruency]

(ii) Since, ∆ABE ≅∆ACF
∴AB = AC [By C.P.C.T.]
⇒ABC is an isosceles triangle.

Question 5. ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.

Solution:
In ∆ABC, we have
AB = AC [ABC is an isosceles triangle]
∴∠ABC = ∠ACB …(1)
[Angles opposite to equal sides of a ∆ are equal]
Again, in ∆BDC, we have
BD = CD [BDC is an isosceles triangle]
∴∠CBD = ∠BCD …(2)
[Angles opposite to equal sides of a A are equal]
Adding (1) and (2), we have
∠ABC + ∠CBD = ∠ACB + ∠BCD
⇒∠ABD = ∠ACD.

Question 6. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

Solution:
AB = AC [Given] …(1)
From (1) and (2), we have
Now, in ∆ABC, we have
∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]
⇒2∠ACB + ∠BAC = 180° …(3)
[∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)]
Similarly, in ∆ACD,
⇒2∠ACD + ∠CAD = 180° …(4)
[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°
⇒2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]
⇒2∠BCD = 360° – 180° = 180°
⇒∠BCD = = 90°
Thus, ∠BCD = 90°

Question 7. ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.

Solution:
In ∆ABC, we have AB = AC [Given]
∴Their opposite angles are equal.

⇒∠ACB = ∠ABC
Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]
⇒90° + ∠B + ∠C = 180° [∠A = 90°(Given)]
⇒∠B + ∠C= 180°- 90° = 90°
But ∠B = ∠C
∠B = ∠C = = 45°
Thus, ∠B = 45° and ∠C = 45°

Question 8. Show that the angles of an equilateral triangle are 60° each.

Solution:
In ∆ABC, we have

AB = BC = CA
[ABC is an equilateral triangle]
AB = BC
⇒∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]
Similarly, AC = BC
⇒∠A = ∠B …(2)
From (1) and (2), we have
∠A = ∠B = ∠C = x (say)
Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]
∴x + x + x = 180o
⇒3x = 180°
⇒x = 60°
∴∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

### NCERT Solutions for Class 9 Maths Exercise 7.3

Question 1. ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (See figure). If AD is extended to intersect BC at P, show that:

(i) ABD ACD

(ii) ABP ACP

(iii) AP bisects A as well as D.

(iv) AP is the perpendicular bisector of BC.

Solution:
(i)ABC is an isosceles triangle.

AB = AC

DBC is an isosceles triangle.

BD = CD

Now in ABD and ACD,

AB = AC [Given]

BD = CD [Given]

ABD ACD [By SSS congruency]

(ii)Now in ABP and ACP,

AB = AC [Given]

AP = AP

ABP ACP [By SAS congruency]

(iii)Since ABP ACP [From part (ii)]

BAP = CAP [By C.P.C.T.]

AP bisects A.

Since ABD ACD [From part (i)]

Now ADB + BDP = [Linear pair] ……….(iii)

And ADC + CDP = [Linear pair] ……….(iv)

From eq. (iii) and (iv),

BDP = CDP

DP bisects D or AP bisects D.

(iv)Since ABP ACP [From part (ii)]

BP = PC [By C.P.C.T.] ……….(v)

And APB = APC [By C.P.C.T.] …….(vi)

Now APB + APC = [Linear pair]

APB + APC = [Using eq. (vi)]

2APB =

APB =

AP BC ……….(vii)

From eq. (v), we have BP PC and from (vii), we have proved AP B. So, collectively AP is perpendicular bisector of BC.

Question 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:

Solution:
In ABD and ACD,

AB = AC [Given]

ABD ACD [RHS rule of congruency]

BD = DC [By C.P.C.T.]

Question 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that
(i) ∆ABC ≅∆PQR
(ii) ∆ABM ≅∆PQN

Solution:
In ∆ABC, AM is the median.
∴BM = BC ……(1)
In ∆PQR, PN is the median.
∴QN = QR …(2)
And BC = QR [Given]
⇒BC = QR
⇒BM = QN …(3) [From (1) and (2)]

(i) In ∆ABM and ∆PQN, we have
AB = PQ , [Given]
AM = PN [Given]
BM = QN [From (3)]
∴∆ABM ≅∆PQN [By SSS congruency]

(ii) Since ∆ABM ≅∆PQN
⇒∠B = ∠Q …(4) [By C.P.C.T.]
Now, in ∆ABC and ∆PQR, we have
∠B = ∠Q [From (4)]
AB = PQ [Given]
BC = QR [Given]
∴∆ABC ≅∆PQR [By SAS congruency]

Question 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:
Since BE ⊥AC [Given]

∴BEC is a right triangle such that ∠BEC = 90°
Similarly, ∠CFB = 90°
Now, in right ∆BEC and ∆CFB, we have
BE = CF [Given]
BC = CB [Common hypotenuse]
∠BEC = ∠CFB [Each 90°]
∴∆BEC ≅∆CFB [By RHS congruency]
So, ∠BCE = ∠CBF [By C.P.C.T.]
or ∠BCA = ∠CBA
Now, in ∆ABC, ∠BCA = ∠CBA
⇒AB = AC [Sides opposite to equal angles of a ∆ are equal]
∴ABC is an isosceles triangle.

Question 5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥BC to show that ∠B = ∠C.

Solution:
We have, AP ⊥BC [Given]

∠APB = 90° and ∠APC = 90°
In ∆ABP and ∆ACP, we have
∠APB = ∠APC [Each 90°]
AB = AC [Given]
AP = AP [Common]
∴∆ABP ≅∆ACP [By RHS congruency]
So, ∠B = ∠C [By C.P.C.T.]

### NCERT Solutions for Class 9 Maths Exercise 7.4

Question 1. Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:
Let us consider ∆ABC such that ∠B = 90°
∴∠A + ∠B + ∠C = 180°
⇒∠A + 90°-+ ∠C = 180°
⇒∠A + ∠C = 90°
⇒∠A + ∠C = ∠B
∴∠B > ∠A and ∠B > ∠C

⇒Side opposite to ∠B is longer than the side opposite to ∠A
i.e., AC > BC.
Similarly, AC > AB.
Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side.

Question 2. In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:
∠ABC + ∠PBC = 180° [Linear pair]
and ∠ACB + ∠QCB = 180° [Linear pair]
But ∠PBC < ∠QCB [Given] ⇒180° – ∠PBC > 180° – ∠QCB
⇒∠ABC > ∠ACB
The side opposite to ∠ABC > the side opposite to ∠ACB
⇒AC > AB.

Question 3. In figure, ∠B <∠A and ∠C <∠D. Show that AD < BC.

Solution:
Since ∠A > ∠B [Given]
∴OB > OA …(1)
[Side opposite to greater angle is longer]
Similarly, OC > OD …(2)
Adding (1) and (2), we have
OB + OC > OA + OD

Question 4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B >∠D.

Solution:
Let us join AC.

Now, in ∆ABC, AB < BC [∵AB is the smallest side of the quadrilateral ABCD] ⇒BC > AB
⇒∠BAC > ∠BCA …(1)
[Angle opposite to longer side of A is greater]
Again, in ∆ACD, CD > AD
[ CD is the longest side of the quadrilateral ABCD]
[Angle opposite to longer side of ∆ is greater]
Adding (1) and (2), we get
∠BAC + ∠CAD > ∠BCA + ∠ACD
⇒∠A > ∠C
Similarly, by joining BD, we have ∠B > ∠D.

Question 5. In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ.

Solution:
In ∆PQR, PS bisects ∠QPR [Given]
∴∠QPS = ∠RPS
and PR > PQ [Given]
⇒∠PQS > ∠PRS [Angle opposite to longer side of A is greater]
⇒∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS]
∵Exterior ∠PSR = [∠PQS + ∠QPS]
and exterior ∠PSQ = [∠PRS + ∠RPS]
[An exterior angle is equal to the sum of interior opposite angles]
Now, from (1), we have
∠PSR = ∠PSQ.

Question 6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Let us consider the ∆PMN such that ∠M = 90°

Since, ∠M + ∠N+ ∠P = 180°
[Sum of angles of a triangle is 180°]
∵∠M = 90° [PM ⊥l]
So, ∠N + ∠P = ∠M
⇒∠N < ∠M
⇒PM < PN …(1)
Similarly, PM < PN1…(2)
and PM < PN2…(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular line segment is the shortest line segment drawn on a line from a point not on it.

### NCERT Solutions for Class 9 Maths Exercise 7.5

Question 1. ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Solution:
Let us consider a ∆ABC.
Draw l, the perpendicular bisector of AB.
Draw m, the perpendicular bisector of BC.
Let the two perpendicular bisectors l and m meet at O.
O is the required point which is equidistant from A, B and C.

Note: If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle.

Question 2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Let us consider a ∆ABC.

Draw m, the bisector of ∠C.
Let the two bisectors l and m meet at O.
Thus, O is the required point which is equidistant from the sides of ∆ABC.
Note: If we draw OM ⊥BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle. Point O is called incentre of the triangle.

Question 3. In a huge park, people are concentrated at three points (see figure)

A: where these are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exist.
Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?
[Hint The parlour should be equidistant from A, B and C.]

Solution:
Let us join A and B, and draw l, the perpendicular bisector of AB.
Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O.
The point O is the required point where the ice cream parlour be set up.
Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O.

Question 4. Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Solution:
It is an activity.
We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii).
∴The Fig. (ii) has more triangles.

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