Class 9 Maths Chapter 7 Triangles

Exercise 7.1

 Question 1.
In a quadrilateral ACBD, AC = AD, and AB bisects ∠A (see figure). Show that Δ ABC ≅ ΔABD. What can you say about and BD?
Solution:

Given : In quadrilateral ACBD.
AC = AD and ∠BAC = ∠DAB
To prove : ΔABC = ΔABD
Proof: In ΔABC and ΔABD, AC = AD [given]
∠BAC = ∠DAB                 [given]
and   B = AB                    [common side]
Hence, ΔABC = ΔABD  [by SAS congrunce rule]
Then, BC = BD
Thus, BC and BD are equal.

 Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that
(i) Δ ABD ≅ Δ BAC
(ii) BD = AC
(iii) Δ ABD = Δ BAC
Solution:

Given : In quadrilateral ABCD.
AD = BC and ∠DAB = ∠CBA
To prove :
(i) ΔABD ≅ ΔBAC
(ii) BD= AC
(iii) ∠ABD = ∠BAC
Proof : (i) In ΔABD and ΔBAC,
AD-BC     [given]
ΔDAB = ΔCBA    [given]
and AB = AB  [common side]
∴ Δ ABD ≅ Δ BAC [by SAS congruence rule]

(ii) From part (i), Δ ABD ≅ ΔBAC
Then, BD = AC [by CPCT]

(iii) From part (i), Δ ABD ≅ Δ BAC
Then, ∠ABD = ∠BAC [by CPCT]
Hence proved.

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

Solution:
Given: AD ⊥ AB and BC ⊥ AB
To prove: CD bisects AB.
Proof: In Δ AOD and Δ BOC.
AD = BC  [given]
∠OAD = ∠OBC  [each 90°]
and  ∠AOD =∠BOC [vertically opposite angles]
∠AOD = ∠BOC [by AAS congruence rule]
Then,OA = OB  [by CPCT]
Thus, CD bisects AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that
ΔABC = ΔCDA.

Solution:
Given : l || m and p || q
To prove : ΔABC = ΔCDA
Proof : In ΔABC and ΔADC,
∠BAC = ∠ACD [alternative interior angles as p || q]
∠ACB = ∠DAC [alternative interior angles as l || m]
AC = AC [common side]
ΔABC ≅ ΔCDA  [by AAS congruence rule]

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.). Show that
(i) ΔAPB = ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A

Solution:
Consider ΔAPB and ΔAQB, we have
∠APB = ∠AQB =90°
∠PAB = ∠QAB  [∵ AB bisects ∠PAQ]
AB = AB      [common]
∴ By AAS congruence axiom, we have
ΔAPB ≅  ΔAQB, which proves (i)
=>   BP = BQ   [by CPCT]
i.e., B is equidistant from the arms of ∠A, which proves (ii).

Question 6.
In figure, AC = AE, AB – AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution:

In Δ ABC and Δ ADE,
We have  AB = AD (Given)
∠BAD = ∠EAC (Given) …(i)
On adding ∠DAC on both sides in Eq. (i)
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
and    AC – AE  [given]
Δ ABC ≅ Δ ADE [by AAS congruence rule]
BC = DE  [by CPCT]

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 figure). Show that
(i) Δ DAP ≅ ΔEBP
(ii) AD = BE

Solution:
We have AP = BP    [∵ P is the mid-point of AB (Given) ]…(i)
∠EPA = ∠DPB  (Given) …. (ii)
∠BAD = ∠ABE  (Given) …(iii)
On adding ∠EPD on both sides in Eq. (ii),
we have => ∠EPA + ∠EPD = ∠DPB + ∠EPD
=>  ∠DPA = ∠EPB   …(iv)
Now, in Δ DAP and Δ EBP, we have …..(iv)
∠DPA = ∠EPB  [from eq.(iv)]
∠DAP = ∠EBP  (Given)
and  AP = BP [from eq.(i)]
ΔDAP ≅ ΔEBP [by ASA congruence axiom]
Hence,AD = BE [by CPCT]

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 fig.). Show that:
(i) Δ AMC = Δ BMD
(ii) ∠DBC is a right angle
(iii) Δ DBC = Δ ACB
(iv) CM=AB

Solution:
Given: Δ ACB in which ZC = 90° and M is the mid-point of AB.
To Prove :
(i) Δ AMC ≅ Δ BMD
(ii) ∠DBC is a right angle
(iii) Δ DBC ≅ ΔACB
(iv) CM =   AB
Proof : Consider Δ AMC and Δ BMD,
we have AM – BM
CM = DM
∠AMC = ∠BMD
∴ Δ AMC = Δ BMD
AC = DB
and  ∠1–∠2
But ∠1 and ∠2 are alternate angles.
=> BD || CA
Now, BD || CA and BC is transversal.
∴ ∠ACB + ∠CBD = 180° =>90°+ ∠CBD = 180° => ∠CBD =90°
In ΔDBC and ΔACB, we have
CB = BC [common]
DB = AC [using eq.(i)]
∠CBD = ∠BCA  [each= 90°]
ΔDBC ≅ = ΔACB [by ASA congruence axiom]
=> DB = AB =>  DC =>  AB=CM
=> CM =  [∵ CM =  DC]

Exercise 7.2

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

(i) In ΔABC, we have AB = AC   (Given)
=> ∠B = ∠C
∠B = ∠C
=> ∠OBC    =   ∠OCB
and  ∠OBA = ∠OCA
(∵ OB and OC are bisectors of ∠B and ∠C respectively).
(∵ ∠OBC = ∠B and ∠OCB = ∠C)
=> OB   – OC (∵ Sides opposite to equal angles are equal)

(ii) In Δ ABO and ΔACD, we have
AB = AC (Given)
∠OBA = ∠OCA  [From Eq. (i)]
=>  OB = OC    [From Eq. (ii)]
ΔABO ≅ ΔACO  (By SAS congruence axiom)
=>  ∠BAO    =  ∠CAO     (By CPCT)
=> AO is the bisector of ∠BAC.

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

Solution:
Given : AD is the perpendicular bisector of BC;
To prove : In Δ ABC is an isosceles triangle.  i.e.,
AB = AC
Proof : In Δ ADB and Δ ADC,
AD = AD [common]
BD = DC (Given)
and  ∠ADB = ∠ADC  [each= 90°]
Δ ADB ≅ Δ ADC (By SAS congruence axiom)
=>  AB = AC  (By CPCT)
So, Δ 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:
Given: ΔABC is an isosceles triangle in which
AB = AC,BE ⊥ AC and CF ⊥  AB
To prove : BE = CF
Proof : In Δ ABE and Δ ACE,
AB – AC   given]
∠AEB = ∠AFC      [each 90 °, BE ⊥ AC and CF ⊥ AB]
∠BAE = ∠CAF   [common angle]
 ∴ Δ ABE ≅ Δ ACF   [by AAS congruence rule]
Then BE = CF   [by CPCT] Hence proved.

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

(i) ΔABE = ΔACF
(ii) AB = AC
i.e., ΔABC is an isosceles triangle.
Solution:
Given: ΔABC in which BE ⊥ AC and CF⊥  AB, such that BE = CF.
To prove : (i) Δ ABE ≅ Δ ACF
(ii) AB = AC
Proof: (i) In Δ ABC and Δ ACF
∠AEB= ∠AFC   [each 90 °]
∠BAE = ∠CAF [common angle]
and BE = CF  (Given)
Δ ABE ≅ Δ ACF  [by AAS congruence rule]

(ii) From part (i),  Δ ABE ≅ Δ ACF
AB = AC  [by CPCT]

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

Solution:
Consider Δ ABC, we have
AB= AC
=> ∠ABC =    ∠ACB  …(i) [∵ angles opposite to equal sides are equal]
Consider ADBC , we have
BD = CD
=> ∠DBC =    ∠DCB  …(ii) [∵ angles opposite to equal sides are equal]
Adding eqs. (i) and (ii), we get
∠ABC + ∠DBC =  ∠ACB + ∠DCB
=> ∠ABD =    ∠ACD

Question 6.
Δ 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:

In ΔABC, we have AB = AC (Given)
=> ∠ACB = ∠ABC ………..(i) (∵Angles opposite to  equal sides are equal)
Now, AB = AD (Given)
∴ AC = AD (∵ AB = AC)
Now, in ∠ADC, we have
AD= AC (from above)
=> ∠ACD = ∠ADC ……..(ii) (∵ Angles opposite to equal sides are equal)
On adding eqs. (i) and (ii), we have
∠ACB + ∠ACD = ∠ABC + ∠ADC
=»∠BCD = ∠ABC + ∠BDC (∵  ∠ADC = BDC)
Adding ∠BCD on both sides, we have
∠BCD + ∠BCD = ∠ABC + ∠BDC + ∠BCD
=>  2∠BCD = 180° (By Δ property)
∠BCD = 90°

Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB =  AC. Find ∠B and ∠C.
Solution:
Given, Δ ABC is a right angled triangle in which ∠A = 90° and AB = AC.
Then, ∠C = ∠B  (i) [since, angles opposite to equal sides of a triangle are equal]
Now, ∠A + ∠B + ∠C = 180°[since, sum of three angles of a triangle is 180°]
=> 90 °+∠B + ∠B = 180°  [from eq. (i)]
=> 2∠B =90°  ∴ ∠B =45°
Hence, ∠B = 45° and ∠C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Given: Δ ABC is an equilateral triangle.

To prove :  ∠A = ∠B = ∠C = 60°
Proof : Since, all three angles of an equilateral triangle are equal. i.e., ∠A=∠B=∠C    …(i)
∴ Z∠A + ∠B + ∠C = 180° [since, sum of three angles of a triangle is 180°]
=>  ∠A + ∠A + ∠A = 180°  [from eq. (i)]
3∠A = 180°
∠A 
=>∠A = 60°
Hence, ∠A = ∠B = ∠C = 60°
Hence proved

 

Exercise 7.3

Question 1.
Δ ABC and Δ 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 ΔD is extended to intersect BC at P, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector g of

Solution:
Given: Δ ABC and Δ DBC are two isosceles triangles having common base BC, such that AB = AC and DB = DC.
To prove : (i) Δ ABD ≅ Δ ACD
(ii) ΔABP ≅ Δ ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC


Question 2.
AD is an altitude of an isosceles ∠ABC in which AB = AC. Show that
(i) AD bisects BC.
(ii) AD bisects ∠A.
Solution:
Given: Δ ABC is an isosceles triangle and AD⊥ BC, AB = AC.

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see figure). Show that
(i) ΔABM ≅ ΔPQN
(ii)  Δ ABC ≅ ΔPQR

Solution:

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:



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

 

Exercise 7.4

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

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

Question 3.
In figure ∠B <∠A and ∠C<∠D.show that AD<BC
Solution:

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see fig.). Show that ∠A > ∠C and ∠B > ∠D.
Solution:
Given: ABCD is a quadrilateral. AB is the shortest side and CD is the longest side.
To Prove : ∠A> ∠C and ∠B > ∠D.
Construction: Join A and C, also B and D.
Proof: In Δ ABC, AB is the smallest side.
BC > AB
∠1 > ∠2   …(i)
[∵ angle opposite to longer side greater]

Question 5.
In the figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.
Solution:
In Δ PQR, we have

Question 6.
Show that of all line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.
Solution:

Given: x is a line and A is a point not lying on x. AB ⊥ x, C is any point on x other than B.
To prove :  AB < AC
Proof: In ΔABC, ∠B is the right angle.
∴ ∠C is an acute angle.
∴ ∠B > ∠C
=> AC >AB
(∵ the Side opposite to greater angle is longer)
=>AB< AC
Hence, the perpendicular line segment is the shortest.

 

Exercise 7.5

Question 1.
ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of Δ ABC.
Solution:

Let OD and OE be the perpendicular bisectors of sides BC and CA of Δ ABC.
∴ 0 is equidistant from two ends B and C of line-segment BC as 0 lies on the perpendicular bisector of BC.
Similarly, 0 is equidistant from C and A. Thus, the point of intersection 0 of the perpendicular bisectors of sides BC, CA and AB is the required point.

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Let BE and CF be the bisectors of ∠ABC and ∠ACB respectively intersecting AC and AB at E and F respectively.

Since 0 lies on BE, the bisector of ∠ABC, hence O will be equidistant from AB and BC. Again, O lies on the bisector CF of ∠ACB. Hence, 0 will be equidistant from BC and AC. Thus, 0 will be equidistant from AB, BC, and CA.

Question 3.
In a huge park, people are concentrated at three points (see figure)
A.where there are different sides and swings for children
B. near which a man-made lake is situated
C. which is near to a large parking and exit.
Where should an ice-cream parlor be set up so that maximum number of persons can approach it?
[Hint: The parlor should be equidistant from A, B and C.]
Solution:
The ice-cream parlor should be equidistant from A, B, and C for which the point of intersection of perpendicular bisectors of AB, BC, and CA should be situated.
So, 0 is the required point which is equidistant from A, B, and C.

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

Solution:
We first divide the hexagon into six equilateral triangles of side 5 cm as follows:

We take one triangle from six equilateral triangles as shown above and ‘ make as many equilateral triangles of one side 1 cm as shown in the figure given below.

The number of equilateral triangles of side 1cm = 1+ 3 + 5 + 7 + 9 = 25
So, the total number of triangles in the hexagon = 6 x 25 = 150.
To find the number of triangles in Fig. (ii), we adopt the same procedure.
So, the number of triangles in Fig. (ii) = 12 x 25 = 300 Hence, Fig. (ii) has more triangles.