SELINA Solution Class 9 Chapter 10 Isosceles Triangles Exercise 10A

Question 1

In the figure alongside,

AB = AC
∠A = 48°  and
∠ACD = 18° 
Show that BC = CD.

Sol:

In ΔABC,
∠BAC + ∠ACB + ∠ABC = 180°
48° + ∠ACB + ∠ABC = 180°
But ∠ACB = ∠ABC ........[ AB = AC ]
2∠ABC = 180° - 48°
2∠ABC = 132°
∠ABC = 66° = ∠ACB ..........(i)

∠ACB = 66°
∠ACD + ∠DCB = 66°
18° + ∠DCB = 66°
∠DCB = 48° ..........(ii)

Now, In ΔDCB,
∠DBC = 66° .......[ From (i), Since ∠ABC = ∠DBC ]
∠DCB = 48° .......[From (ii)]
∠BDC = 180°- 48° - 66°
∠BDC = 66°
Since ∠BDC = ∠DBC
Therefore, BC = CD
Equal angles have equal sides opposite to them.

Question 2

Calculate:
(i) ∠ADC
(ii) ∠ABC
(iii) ∠BAC

Sol:

Given: ACE = 130°; AD = BD = CD

Proof:
(i) ∠ACD + ∠ACE = 18° ....... [ DCE is a st. line ]
⇒ ∠ACD = 180° − 130°
⇒ ∠ACD = 50° 
Now, CD = AD
⇒ ∠ACD = ∠DAC = 50°  ..... (i)[ Since angels opposite to equal sides are equal]

In ΔADC,
∠ACD = ∠DAC = 50°  
∠ACD + ∠DAC + ∠ADC = 180° 
 50°  + 50°  + ∠ADC = 180° 
∠ADC =180° − 100°
∠ADC = 80°

(ii) ∠ADC = ∠ABD + ∠DAB ....[Exterior angle is equal to sum of opp. interor angle] 
But AD = BD
∴ ∠DAB = ∠ABD
⇒ 80° = ∠ABD + ∠ABD
⇒ 2∠BD = 80°
⇒ ∠ABD = 40° = ∠DAB .....(ii)

(iii) ∠BAC = ∠DAB + ∠DAC
substituting the value from (i) and (ii)
∠BAC = 40° + 50°
⇒ ∠BAC = 90°

Question 3

In the following figure, AB = AC; BC = CD and DE are parallel to BC.
Calculate:
(i) ∠CDE
(ii) ∠DCE

Sol:

∠FAB = 128° .......[ Given ]
∠BAC + ∠FAB = 180° ......[ FAC is a st. line ] 
⇒ ∠BAC = 180° − 128° 
⇒ ∠BAC = 52° 

In ΔABC,
∠A = 52°
∠B=  ∠C .....[Given AB = AC and angels opposite to equal sides are equal]

∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠B + ∠B = 180° 
⇒ 52° + 2∠B = 180°
⇒ 2∠B = 128°
⇒ ∠B = 64° = ∠C ........(i)
⇒ ∠B = ∠ADE .......[ Given DE || BC ]
(i) 
Now,
∠ADE + ∠CDE + ∠B = 180° ....[ ADB is a st. line ]
⇒  64° + ∠CDE + 64° = 180°
⇒ ∠CDE = 180° − 128° 
⇒ ∠CDE = 52° 

(ii)
Given DE || BC and DC is the transversal.
⇒ ∠CDE = ∠DCB = 52° .......(ii)
Also, ∠ECB = 64° .....[ From (i) ]
But,
∠ECB = ∠DCE + ∠DCB
⇒ 64° = ∠DCE + 52°
⇒ ∠DCE + 64° − 52°
⇒  ∠DCE = 12°

Question 4.1

Calculate x :

Sol:

Let the triangle be ABC and the altitude be AD.

In ΔABD,
∠DBA = ∠DAB = 37° .....[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB .......[Exterior angle is equal to the sum of opp. interior angles]
∴ ∠CDA = 37° + 37° 
⇒ ∠CDA = 74°
Now in ΔADC,
∠CDA = ∠CAD = 74° ....[ Given CD = AC and angles opposite to equal sides are equal]
Now,
∠CAD + ∠CDA + ∠ACD = 180°
⇒ 74° + 74° + x = 180° 
⇒ x = 180° − 148° 
⇒ x = 32°

Question 4.2

Calculate x :

Sol:

Let triangle be ABC and altitude be AD.

In ΔABD,
∠DBA = ∠DAB = 50° ...[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB ...[Exterior angle is equal to the sum of opp. interior angles]
∴ ∠CDA = 50° + 50° 
⇒ ∠CDA = 100°

In ΔADC,
∠DAC = ∠DCA = x  ...[Given AD = DC and angles opposite to equal sides are equal]
∴ ∠DAC + ∠DCA + ∠ADC = 180°
⇒ x + x + 100° = 180°
⇒ 2x = 80°
⇒ x = 40°

Question 5

In the figure, given below, AB = AC.
Prove that: ∠BOC = ∠ACD.

Sol:

Let  ∠ABO = ∠OBC = x and ∠ACO = ∠OCB = y
In ΔABC,
∠BAC = 180° - 2x - 2y          ....(i)
Since, ∠B = ∠C                     ...[AB = AC]
${\displaystyle \frac{1}{2}\text{B}=\frac{1}{2}\text{C}}$
⇒ x = y
Now, 
∠ACD = 2x + ∠BAC        ...[ Exterior angle is equal to sum of opp. interior angles ]
∠ACD = 2x + 180° - 2x - 2y      ...[ From(i) ] 
∠ACD =180° - 2y                      ....(i)

In ΔOBC,
∠BOC = 180°- x - y 
⇒ ∠BOC = 180°- y - y             ...[ Already proved ]
⇒ ∠BOC = 180° - 2y                ...(ii)

From (i) and (ii)
∠BOC = ∠ACD

Question 6

In the figure given below, LM = LN; angle PLN = 110^o.

calculate: (i) ∠LMN
                 (ii) ∠MLN

Sol:

Given: ∠PLN = 110°
(i) We know that the sum of the measure of all the angles of a quadrilateral is 360°. 
In quad. PQNL,
∠QPL + ∠PLN + ∠LNQ + ∠NQP = 360°
⇒ 90° + 110° + ∠LNQ + 90° = 360°
⇒ ∠LNQ = 360° − 290°
⇒ ∠LNQ = 70°
⇒ ∠LNM = 70° ........(i)
In ΔLMN,
LM = LN ........( Given )
∴ ∠LNM = ∠LMN ....... [angles opp. to equal sides are equal]
⇒ ∠LMN = 70° ....(ii) [ from(i) ]

(ii) In ΔLMN,
∠LMN + ∠LNM+ ∠MLN = 180°
But ∠LNM= ∠LMN = 70° .....[ From(i) and (ii)]
∴ 70° + 70° + MLN = 180°
⇒ ∠MLN = 180°− 140°
⇒ ∠MLN = 40°

Question 7

An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠ CAB = 55^o.
Find:
(i) ∠DCB 
(ii) ∠CBD.

Sol:

ln ΔABC,
AC = BC .......[Given]
∴ ∠CAB = ∠CBD ........[angles opp.to equal sides are equal]
⇒ ∠CBD = 55°

In ΔABC,
∠CBA + ∠CAB + ∠ACB = 180°
but, ∠CAB = ∠CBA = 55°
⇒ 55° + 55° + ∠ACB = 180°
⇒  ∠ACB = 180° − 110°
⇒  ∠ACB = 70°
Now,
In ΔACD and ΔBCD,
AC = BC .......[Given]
CD = CD ........[Common]
AD = BD .........[Given: CD bisects AB]
∴ ΔACD ≅ ΔBCD
⇒ ∠DCA = ∠DCB
⇒ ∠DCB = ${\displaystyle \frac{\angle \text{ACB}}{2}=\frac{70°}{2}}$
⇒ ∠DCB = 35°.

Question 8

Find x :

Sol:

Let us name the figure as following :

In ΔABC, 
AD = AC .......[ Given ] 
∴ ∠ADC =∠ACD ......[Angles opp. to equal sides are equal]
⇒ ∠ADC = 42°

Now, 
∠ADC = ∠DAB +∠DBA ...[Exterior angle is equal to the sum of opp. interior angles]
But, 
∠DAB = ∠DBA ......[Given: BD = DA]
∴ ∠ADC = 2∠DBA
⇒ 2∠DBA = 42°
⇒ ∠DBA = 21°

For x:
x = ∠CBA + ∠BCA .......[Exterior angle is equal to the sum of opp. interior angles]
We know that,
∠CBA = 21°
∠BCA = 42°
∴ x = 21 + 42°
⇒ x = 63°

Question 9

In the triangle ABC, BD bisects angle B and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, find the values of x and y.

Sol:

In ΔABD aand ΔDBC,
BD  = BD                 ...[ Common ]
∠BDA = ∠BDC        ...[ each equal to 90° ]
∠ABD = ∠DBC        ...[ BD bisects ∠ABC ]
∴ ΔABD ≅ ΔDBC       ...[ ASA criterion ]

Therefore,
AD = DC
x + 1 = y + 2
⇒ x = y + 1               ..... (i)
and AB = BC
3x + 1 = 5y - 2

Subtituting the value of x from (i)
3( y + 1 ) + 1 = 5y - 2 
⇒ 3y + 3 + 1 = 5y - 2
⇒ 3y +  4 = 5y - 2
⇒ 2y = 6
⇒ y = 3

Putting y = 3 in (i)
x = 3 + 1 
∴ x = 4

Question 10

In the given figure; AE || BD, AC || ED and AB = AC. Find ∠a, ∠b and ∠c.

Sol:

Let P and Q be the points as shown below: 

Given:
∠PDQ = 58°
∠PDQ = ∠EDC = 58° .....[ Vertically opp . angles ]
∠EDC = ∠ACB =58° ........[ Corresponding angles  ∵ AC || ED ]

In  ΔABC,
AB = AC .......[ Given ]
∴ ∠ACB = ∠ABC = 58° ......[angels opp. to equal sides are equal]
Now, 
∠ACB + ∠ABC+ ∠BAC = 180°
⇒ 58° + 58° + a = 180°
⇒ a = 180° − 116°
⇒ a = 64°
Since AE || BD and AC is the transversal. 
∠ABC = b .......[ Corresponding angles ]
∴ b = 58°

Also since AE || BD and ED is the transversal 
∠EDC = c  .......[ Corresponding angles ] 
∴ c = 58°

Question 11

In the following figure; AC = CD, AD = BD and ∠C = 58^o.

Find the angle CAB.

SOl:

In ΔACD,
AC = CD                      ...[ Given]
∴ ∠CAD = ∠CDA
∠ACD = 58°                ...[ Given ]

∠ACD + ∠CDA + ∠CAD = 180°
⇒ 58° + 2∠CAD = 180°
⇒ 2∠CAD = 122°
⇒ ∠CAD = ∠CDA = 61° ...(i) 

Now,
∠CDA = ∠DAB + ∠DBA ...[ Ext. angel is equal to sum of opp. int. angles ]
But,
∠DAB = ∠DBA                ...[ Given : AD = DB ]
∴ ∠DAB +∠ DAB = ∠CDA
⇒ 2∠DAB = 61°
⇒ ∠DAB = 30.5°             ....(ii)

In ΔABC,
∠CAB = ∠CAD +∠DAB
∴ ∠CAB = 61° + 30.5°
⇒ ∠AB = 91.5°

Question 12

In the figure given below, if AC = AD = CD = BD; find angle ABC.

Sol:

In ΔACD,
AC = AD = CD ......[Given]
Hence, ACD is an equilateral triangle.
∴ ∠ACD = ∠CDA = ∠CAD = 60°
∠CDA = ∠DAB + ∠ABD ........[Ext angle is equal to the sum of opp. int. angles]
But,
∠DAB = ∠ABD ......[Given: AD = DB]
∴ ∠ABD + ∠ABD = ∠CDA
⇒ 2∠ABD = 60°
⇒ ∠ABD = ∠ABC = 30°

Question 13

In triangle ABC; AB = AC and ∠A : ∠B = 8 : 5; find angle A.

Sol:

Let ∠A = 8x and ∠B = 5x
Given: AB = AC
⇒ ∠B = ∠C = 5x     ...[Angles opp. to equal sides are equal]

Now,
∠A + ∠B + ∠C = 180°
⇒ 8x + 5x + 5x = 180°
⇒ 18x = 180°
⇒ x = 10°

Given that :
∠A = 8x
⇒ ∠A = 8 x 10°
⇒ ∠A = 80°.

Question 14

In triangle ABC; ∠A = 60^o, ∠C = 40^o, and the bisector of angle ABC meets side AC at point P. Show that BP = CP.

SOl:

In ΔABC,
∠A = 60°
∠C = 40°
∴ ∠B = 180° - 60° - 40°
⇒ ∠B = 80°

Now,
BP is the bisector of ∠ABC.

∴ ∠PBC = ${\displaystyle \frac{\text{∠ABC}}{2}}$

⇒ ∠PBC = 40°
In ΔPBC,
∠PBC = ∠PCB = 40°
∴ BP = CP             ....[ Sides opp. to equal angles are equal.]

Question 15

In triangle ABC; angle ABC = 90^o and P is a point on AC such that ∠PBC = ∠PCB.
Show that: PA = PB.

Sol:

Let PBC = PCB = x
In the right angled triangle ABC,
∠ABC = 90°
∠ACB = x
⇒ ∠BAC = 180° - ( 90° + x )
⇒  ∠BAC = ( 90°- x )             ...(i)

and
∠ABP = ∠ABC - ∠PBC
⇒  ∠ABP = 90° - x                   ...(ii)

Therefore in the triangle ABP;
 ∠BAP = ∠ABP
Hence, PA = PB     ...[sides opp. to equal angles are equal]

Question 16

ABC is an equilateral triangle. Its side BC is produced up to point E such that C is mid-point of BE. Calculate the measure of angles ACE and AEC.

SOl:

ΔABC is an equilateral triangle.
⇒ Side AB = Side AC
⇒ ∠ABC = ∠ACB ........[If two sides of a triangle are equal, then angles opposite to them are equal]

Similarly, Side AC = Side BC
⇒ ∠CAB = ∠ABC .......[If two sides of a triangle are equal, then angles opposite to them are equal]

Hence, ∠ABC = ∠CAB = ∠ACB  = y(say)
As the sum of all the angles of the triangle is 180°.
∠ABC + ∠CAB + ∠ACB = 180°
⇒ 3y = 180°
⇒ y = 60°

∠ACB = ∠ACB = ∠ABC = 60°
Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.
⇒ ∠CAB + ∠CBA = ∠ACE
⇒  60° + 60° = ∠ACE
⇒  ∠ACE = 120°

Now ΔACE is an isosceles triangle with AC = CF
⇒ ∠EAC = ∠AEC
Sum of all the angles of a triangle is 180°
∠EAC + ∠AEC + ∠ACE = 180°
⇒ 2∠AEC + 120° = 180°
⇒  2∠AEC = 180° − 120°
⇒  ∠AEC = 30°

Question 17

In triangle ABC, D is a point in AB such that AC = CD = DB. If ∠B = 28°, find the angle ACD.

Sol:

ΔDBC is an isosceles triangle.
As, Side CD = Side DB
⇒ ∠DBC = ∠DCB .......[If two sides of a triangle are equal, then angles opposite to them are equal]

And ∠B = ∠DBC = ∠DCB = 28°
As the sum of all the angles of the triangle is 180°
∠DCB + ∠DBC + ∠BCD = 180°
⇒ 28° + 28° + ∠BCD = 180°
⇒ ∠BCD = 180° − 56°
⇒ ∠BCD = 124°
Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.
⇒ ∠DBC+ ∠DCB = ∠DAC
⇒ 28° + 28° = 56°
⇒ DAC = 56°
Now ΔACD is an isosceles triangle with AC = DC
⇒ ∠ADC = ∠DAC = 56°
Sum of all the angles of a triangle is 180°
⇒ ∠ADC + ∠DAC + ∠DCA = 180°
⇒ 56° +56° + ∠DCA = 180°
⇒ ∠DAC = 180° − 112°
⇒ ∠DCA = 64° = ∠ACD.

Question 18

In the given figure, AD = AB = AC, BD is parallel to CA and angle ACB = 65°. Find angle DAC.

SOl:

We can see that the ΔABC is an isosceles triangle with Side AB = Side AC.
⇒ ∠ACB = ∠ABC
As ∠ACB = 65°
hence ∠ABC = 65°
Sum of all the angles of a triangle is 180°
∠ACB + ∠CAB + ∠ABC = 180°
65°+ 65° + ∠CAB = 180°
∠CAB = 180° − 130°
∠CAB = 50°

As BD is parallel to CA
Therefore, ∠CAB = ∠DBA since they are alternate angles.
∠CAB = ∠DBA = 50°
We see that ΔADB is an isosceles triangle with Side AD = Side AB.
⇒ ∠ADB = ∠DBA = 50°
Sum of all the angles of a triangle is 180°
∠ADB + ∠DAB + ∠DBA = 180°
50° + ∠DAB + 50° = 180°
∠DAB = 180° − 100° = 80°
∠DAB = 80°
The angle DAC is the sum of angle DAB and CAB.
∠DAC = ∠CAB + ∠DAB
∠DAC = 50°+ 80°
∠DAC = 130°

Question 19.1

Prove that a triangle ABC is isosceles, if: altitude AD bisects angles BAC.

SOl:

In ΔABC, let the altitude AD bisects ∠BAC.
Then we have to prove that the ΔABC is isosceles.

In triangles ADB and ADC,
∠BAD = ∠CAD    ...(AD is bisector of ∠BAC)
AD = AD             ...(common)
∠ADB = ∠ADC    ....(Each equal to 90°)
⇒ ΔADB ≅ ΔADC ...(by ASA congruence criterion)
⇒ AB = AC           ...(cpct)
Hence, ΔABC is an isosceles.

Question 19.2

Prove that a triangle ABC is isosceles, if: bisector of angle BAC is perpendicular to base BC.

Sol:

In Δ ABC, the bisector of ∠ BAC is perpendicular to the base BC. We have to prove that the ΔABC is isosceles.

In triangles ADB and ADC,
∠BAD = ∠CAD .......(AD is bisector of ∠BAC)
AD = AD ........(common)
∠ADB = ∠ADC .......(Each equal to 90°)
⇒ ΔADB ≅ ΔADC ......(by ASA congruence criterion)
⇒ AB = AC ........(cpct)
Hence, ΔABC is isosceles.

Question 20

In the given figure; AB = BC and AD = EC.
Prove that: BD = BE.

SOl:

In ΔABC,
AB = BC .......(given)
⇒ ∠BCA = ∠BAC  .......(Angles opposite to equal sides are equal)
⇒ ∠BCD = ∠BAE ….(i)
Given, AD = EC
⇒ AD + DE = EC + DE ...(Adding DE on both sides)
⇒ AE = CD .....….(ii)
Now, in triangles ABE and CBD,
AB = BC .....(given)
∠BAE = ∠BCD ....[From (i)]
AE = CD ......[From (ii)]
⇒ ΔABE ≅ ΔCBD
⇒ BE  = BD ....(cpct)

SELINA Solution Class 9 Chapter 9 Triangles (Congruency in Triangles) Exercise 9B

Question 1

On the sides AB and AC of triangle ABC, equilateral triangle ABD and ACE are drawn. 
Prove that:  (i) ∠CAD = ∠BAE 
                   (ii) CD = BE

Sol:

Given:  ΔABD is an equilateral triangle.
ΔACE is an equilateral triangle
We need to prove that 
(i) ∠CAD = ∠BAE 

Proof:
(i) ΔABD is equilateral
∴ Each angel = 60°
⇒ ∠BAD = 60°                            ...(1)
Similarly,
 ΔACE is equilateral
∴ Each angel = 60°
⇒ ∠CAE = 60°                             ...(2)
⇒ ∠BAD = ∠CAE      ...[ from (1) and (2) ]...(3)
Adding ∠BAC to both sides, we have
⇒ ∠BAD + ∠BAC = ∠CAE + ∠BAC
⇒ ∠CAD = ∠BAE                         ...(4)

(ii) In ΔCAD and ΔBAE
AC = AE                 ...[ ΔACE is equilateral ]
∠CAD = ∠BAE      ... [ from (4) ]
AD = AB                ...[ ΔABD is euilateral ]
∴ By Side-Angel-Side criterion of congruency, 
ΔCAD ≅ ΔBAE
The corresponding parts of the congruent
triangles are congruent.
∴ CD = BE             ...[ by c.p.c.t ]
Hence proved.

Question 2.1

In the following diagram, ABCD is a square and APB is an equilateral triangle.

(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC.

Sol:

Given: ABCD is a Square and ΔAPB is an equilateral triangle.
We need to
(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC

(i) Proof:
AP = PB = AB            ...[ APB is an equilateral triangle ]
Also, we have,
∠PBA = ∠PAB = ∠APB = 60°                        ...(1)
Since ABCD is a square, we have
∠A =∠ B = ∠C = ∠D = 90°                        ...(2)
Since ∠DAP = ∠A - PAB                                ...(3)
⇒ ∠DAP = 90° - 60°  
⇒ ∠DAP =30°     ...[ from (1) and (2) ]        ...(4)

Similarly ∠CBP = ∠B - ∠PBA
⇒ ∠CBP = 90°  - 60° 
⇒ ∠CBP = 30°       ...[ from (1) and (2) ] ...(5)
⇒ ∠DAP = ∠CBP  ....[ from (1) and (2) ] ...(6)
In ΔAPD and ΔBPC
AD = BC          ...[ Sides of square ABCD ]
∠DAP = ∠CBP  ...[ from(6) ]
AP= BP           [ Sides of equilateral ΔAPB ]
∴ By Side-Angel-Side Criterion of Congruence, we have,
ΔAPD ≅ ΔBPC

(ii)
AP = PB = AB  ....[ ΔAPB is an equilateral triangle ] ...(7)
AB = BC = CD = DA  ...[ Sides of square ABCD ] ...(8)
From (7) and (8), we have
AP = DA aand PB = BC                                   ... (9)
In ΔAPD,
AP = DA                    ...[ from (9) ]
∠ADP = ∠APD   ...[ Angel opposite to equal sides are equal ]                                        ...(10)
∠ADP + ∠APD+ +∠DAP + 180°  ...[ Sum of angel of a triangle = 180° ]
⇒ ∠ADP + ∠ADP + 30° = 180°    [ from (3), ∠DAP =30° from (10), ∠ADP = ∠APD ]
⇒ ∠ADP + ∠ADP = 180°  -  30°
⇒ 2∠ADP = ${\displaystyle \frac{{150}^0}{2}}$
⇒∠ADP= 75°
We have ∠PDC =∠D - ∠ADP 
⇒∠PDC = 90° - 75°
⇒∠PDC =15°                                                         ...(11)
In BPC, 
PB = BC     ...[ from (9) ]
∴  ∠PCB =∠ BPC  ...[Angel opposite to equal sides are equal ]                                                              ... (12)
∠PCB + ∠BPC + ∠CBP = 180°   ....[ Sum of angel of a triangle = 180°  ]
⇒ ∠PCB + ∠PCB + 30°  = 180°  ....[ from (5), ∠CBP =  from (12) , ∠PCB =∠BPC ]
⇒ 2∠PCB =  180°  -  30°
⇒ 2∠PCB = ${\displaystyle \frac{150°}{2}}$
⇒ ∠PCB = 75°
 We have ∠PCD = ∠C - ∠PCB 
⇒ ∠PCD = 90°  -  75°
⇒ ∠PCD = 15°                                             ... (13)
In ΔDPC, 
∠PDC = 15° 
∠PCD = 15°
∠PCD + ∠PDC + ∠DPC = 180°   ...[ Sum of angles of a triangle = 180°  ]

⇒ 15°  + 15°  + ∠DPC = 180°
⇒ ∠DPC = 180°   -  30°
⇒ ∠DPC = 150° 
∴  Angles of DPC, are:  15° , 150°  , 15° 

Question 2.2

In the following diagram, ABCD is a square and APB is an equilateral triangle.

(i) Prove that: ΔAPD ≅ ΔBPC
(ii) Find the angles of ΔDPC.

Sol:

Given: ABCD is a Square and ΔAPB is an equilateral triangle.

(i) Proof: In ΔAPB,
AP = PB = AB            ...[ APB is an equilateral triangle ]
Also, we have,
∠PBA = ∠PAB = ∠APB = 60°                         ...(1)
Since ABCD is a square, we have
∠A =∠ B = ∠C = ∠D = 90°                           ...(2)
Since ∠DAP = ∠A + ∠PAB                            ..(3)
⇒ ∠DAP = 90° + 60°  
⇒ ∠DAP = 150°     ...[ from (1) and (2) ]        ...(4)

Similarly ∠CBP = ∠B + ∠PBA
⇒ ∠CBP = 90° + 60° 
⇒ ∠CBP = 150°       ...[ from (1) and (2) ] ...(5)
⇒ ∠DAP = ∠CBP  ....[ from (1) and (2) ] ...(6)
In ΔAPD and ΔBPC
AD = BC          ...[ Sides of square ABCD ]
∠DAP = ∠CBP  ...[ from(6) ]
AP= BP           [ Sides of equilateral ΔAPB ]
∴ By Side-AAngel-SIde Criterion of Congruence, we have,
ΔAPD ≅ ΔBPC

(ii)
AP = PB = AB  ....[ ΔAPB is an equilateral triangle ] ...(7)
AB = BC = CD = DA  ...[ Sides of square ABCD ] ...(8)
From (7) and (8), we have
AP = DA aand PB = BC                                  ... (9)
In ΔAPD,
AP = DA                    ...[ from (9) ]
∠ADP = ∠APD   ...[ Angel opposite to equal sides are equal ]                                             ...(10)
∠ADP + ∠APD+ +∠DAP + 180°  ...[ Sum of angel of a triangle = 180° ]
⇒ ∠ADP + ∠ADP + 150° = 180°    [ from (3), ∠DAP =150° from (10), ∠ADP = ∠APD ]
⇒ ∠ADP + ∠ADP = 180° - 150°
⇒ 2∠ADP = 30°
⇒ ∠ADP = ${\displaystyle \frac{{30}^{\circ }}{2}}$
⇒∠ADP= 15°
We have ∠PDC =∠D - ∠ADP 
⇒∠PDC =90° - 15°
⇒∠PDC =75°                                               ...(11)
In ΔBPC, 
PB = BC     ...[ from (9) ]
∴  ∠PCB =∠ BPC  ...[Angel opposite to equal sides are equal ]                                           ... (12)
∠PCB + ∠BPC + ∠CBP = 180°   ....[ Sum of angel of a triangle = 180°  ]
⇒ ∠PCB + ∠PCB + 30°  = 180°  ....[ from (5), ∠CBP = 150° from (12) , ∠PCB =∠BPC ]
⇒ 2∠PCB =  180° - 150°
⇒ 2∠PCB = ${\displaystyle \frac{{30}^{\circ }}{2}}$
⇒ ∠PCB = 15°
 We have ∠PCD = ∠C - ∠PCB 
⇒ ∠PCD = 90° - 15°
⇒ ∠PCD = 75°                                           ... (13)
In ΔDPC, 
∠PDC = 75° 
∠PCD = 75°
∠PCD + ∠PDC + ∠DPC = 180°   ...[ Sum of angles of a triangle = 180°  ]

⇒ 75°  + 75°  + ∠DPC = 180°
⇒ ∠DPC = 180° - 150°
⇒ ∠DPC = 30° 
∴  Angles of DPC, are: 75°, 30° , 75° 

Question 3

In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares.

Prove that: 
(i) ΔACQ and ΔASB are congruent.
(ii) CQ = BS.

Sol:

Given: A(Δ ABC) is right-angled at B.
ABPQ and ACRS are squares

To Prove:
(i) ΔACQ ≅ ΔASB
(ii) CQ = BS

Proof:
(i)
∠ QAB = 90°    ...[ ABPQ is a square ] ...(1)
∠ SAC = 90°     ...[ ACRS is a square ] ...(2)
From (1) and (2) , We have
∠ QAB = ∠SAC                        ...(3)
Adding ∠BAC to both sides of (3), We have
∠ QAB + ∠BAC = ∠SAC + ∠BAC
⇒ ∠QAC = ∠SAB                   ...(4)

In ΔACQ and ΔASB,
QA = QB             ...[ Sides of a square ABPQ ]
∠QAC = ∠SAB    ...[ From(4) ]
AC = AS             ...[ sides of a square ACRS ]
∴ By Side -Angle-Side criterion of congruence,
ΔACQ ≅ ΔASB

(ii) 
The corresponding parts of the congruent triangles are congruent,
∴ CQ = BS           ...[ c.p.c.t. ]

Question 4

In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that: AE is parallel to BC.

Sol:

Given: A(ΔABC) in which BD is the median to AC.
BD is produced to E such that BD = DE,
We need to prove that AE II BC.
Construction: Join AE

Proof:
AD = DC          ...[ BD is median to AC ] ...(1)
In ΔBDC and ΔADE,
BD = DE                      ...[ Given ]
∠BDC = ∠ADE = 90°   ...[ Vertically opposite angles ]
AD = DC                     ...[ from(1) ]
∴ By Side-Angle-Side Criterion of congruence,
ΔBDC ≅ ΔADE
The corresponding parts of the congruent triangles are congruent.
∴ ∠EAD = ∠BCD     ...[ c.p.c.t. ]
But these are alternate angles and AC is the transversal.
Thus, AE || BC.

Question 5

In the adjoining figure, OX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥  PQ ;

prove that: (i) ΔXTQ ≅ ΔXSQ. 
                   (ii) PX bisects angle P.

Sol:

Given: A( ΔPQR ) in which QX is the bisector of ∠Q. and RX is the bisector of ∠R.
XS ⊥ QR and XT ⊥  PQ.
We need to prove that
(i) ΔXTQ ≅ ΔXSQ.
(ii) PX bisects angle P.
Construction: Draw XZ ⊥ PR and join PX.

Proof: 
(i) In ΔXTQ and ΔXSQ,
∠QTX = ∠QSX = 90°    ...[ XS ⊥ QR and XT ⊥  PQ ]
∠TQX = ∠SQX              ...[ QX is bisector of ∠Q ]
QX = QX                      ...[ Common ]
∴ By Angle-Angle-Side Criterion of congruence,
ΔXTQ ≅ ΔXSQ

(ii) The corresponding parts of the congruent triangles are congruent.
∴ XT = XS           ...[ c.p.c.t. ]
In ΔXSR ≅ ΔXZR
∠XSR = ∠XZR = 90°   ...[ XS ⊥ QR and ∠XSR = 90° ]
∠SRX = ∠ZRX             ...[ RX is bisector of ∠R ]
RX = RX                      ....[ Common ]
∴ By Angle-Angle-Side criterion of congruence,
ΔXSR ≅ ΔXZR
The corresponding parts of the congruent triangles are congruent.
∴ XS = XZ             ...[ c.p.c.t. ] ...(2)
From (1) and (2)
XT = XZ                          ....(3)
In ΔXTP and ΔXZP
∠XTP = ∠XZP = 90°       ....[ Given ]
Hyp. XP = Hyp. XP         ....[ Common ]
XT = XZ                         ....[ from(3) ]           
∴ By Right angle-Hypotenuse-side criterion of congruence,
ΔXTP ≅ ΔXZP
The corresponding parts of the congruent triangles are congruent.
∴ ∠XPT = ∠XPZ          ...[ c.p.c.t. ]
∴ PX bisects ∠P.

Question 6

In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA = YC.

SOl:

ABCD is a parallelogram in which ∠A and ∠C are obtuse.

Points X and Y are taken on the diagonal BD.
Such that ∠XAD = ∠YCB = 90°.
We need to prove that XA = YC
Proof:
ln ΔXAD and ΔYCB
∠XAD = ∠YCB= 90°        ...[ Given ]
AD = BC                          ...[ Opposite sides of a parallelogram ]
∠ADX = ∠CBY                 ...[ Alternate angles ]
∴ By Angle-Side-Angle criterion of congruence,
ΔXAD ≅ ΔYCB
The corresponding parts of the congruent triangles are congruent.
∴ XA = YC                    ...[ c.p.c.t. ]
Hence proved.

Question 7

ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively, such produced to E and F respectively, such that AB = BE and AD = DF.
Prove that: ΔBEC ≅ ΔDCF.

Sol:

ABCD is a parallelogram, The sides AB and AD are produced to E and F respectively,
such that AB = BE and AD = DF
We need to prove that ΔBEC ≅ ΔDCF.

Proof: 
AB = DC          ...[ Opposite sides of a parallelogram ] ...(1)
AB = BE           ...[ Given ] ...(2)
From (1) and (2), We have
BF = DC           ...(3)
AD = BC          ...[ Opposite sides of a parallelogram ] ...(4)  
AD = DF          ....[Given]

From (4) and (5), we have
BC = DF                             ...(6)
Since AD II BC, the corresponding angles are equal.
∴ ∠DAB = ∠CBE                ...(7)
Since AB II DC, the corresponding angles are equal.
∴ ∠DAB = ∠FDC                ...(8)
From (7) and (8), we have
∠CBE = ∠FDC

ln ΔBEC and ΔDCF
BF = DC                            ....[ from (3) ]
∠CBE = ∠FDC                   ...[ from (9) ] 
BC = DF                            ....[ from (6) ]
∴ By Side-Angle-Side criterion of congruence,
ΔBEC ≅ ΔDCF
Hence proved.

Question 8

In the following figures, the sides AB and BC and the median AD of triangle ABC are equal to the sides PQ and QR and median PS of the triangle PQR.
Prove that ΔABC and ΔPQR are congruent.

Sol:

Since, BC = QR, We have
BD = QS and DC = SR  ....[ D is the mid-point of BC and S is the mid-point of QR ] 

In ΔABD and ΔPQS,
AB = PQ                  ...(1)
AD = PS                  ...(2)
BD = QS                 ...(3)
Thus, by Side-Side-Side criterion of congruence,
We have ΔABD ≅ ΔPQS
Similarly, in ΔADC and ΔPSR
AD = PS               ...(4)
AC = PR               ...(5)
DC = SR              ....(6)
Thus, by Side-Side-Side criterion of congruence,
We have ΔADC ≅  ΔPSR
We have
BC = BD + DC    ...[ D is the mid-point of BC ]
     = QS + SR     ...[ From (3) and (6) ]
     = QR           ....[ S is the mid-point of QR ] ...(7)
Now consider the triangles ΔABC and ΔPQR
AB = PQ              ...[ from(1) ]
BC = QR              ...[ from(7) ]
AC = PR              ...[ from(7) ]
∴ By Side-Side-Side criterion of congruence, we
have ΔABC ≅  ΔPQR
Hence proved.

Question 9

In the following diagram, AP and BQ are equal and parallel to each other. 

Prove that:  
(i) ΔAOP≅ ΔBOQ.
(ii) AB and PQ bisect each other.

Sol:

In the figure, AP and BQ are equal and parallel to each other. 
∴ AP = BQ and AP || BQ. 
We need to prove that
(i) ΔAOP≅ ΔBOQ.
(ii) AB and PQ bisect each other

(i) ∵ AP || BQ 
∴∠APO =∠BOQ           ...[ Alternate angles ] ...(1)
and ∠PAO =∠QBO       ...[ Alternate angles ] ...(2)
Now in ΔAOP and  ΔBOQ.
∠APO =∠BQO            ...[ from (1) ]
AP = BQ                      ...[ given ]
∠PAO = ∠QBO            ...[ from (1) ]
∴ By Angel-Side-Angel criterion of congruence, we have
ΔAOP≅ ΔBOQ.

(ii) The corresponding parts of the congruent triangles are congruent.
∴ OP = OQ                ...[ c. p. c .t ]
OA = OB                    ...[ c. p. c .t ]
Hence AB and PQ bisect each other.

Question 10

In the following figure, OA = OC and AB = BC.

Prove that:
(i) ∠AOB = 90o
(ii) ΔAOD ≅ ΔCOD
(iii) AD = CD

Sol:

Given:
In the figure, OA=OC, AB =BC
We need to prove that,
AOB = 90° 
(i) In ΔABO and ΔCBO,
AB = BC                   ...[given ]
AO = CO                ...[ given ]
OB = OB               ...[ common ]
∴By Side-Side-Side criterion of congruence, we have
ΔABO ≅ ΔCBO
The corresponding parts of the congruent triangles are congruent.
∴∠ABO = ∠CBO       ...[c. p.c.t. ]
⇒ ∠ABD = ∠CBD        
and ∠AOB = ∠COB   ...[c. p.c t ]
We have
∠AOB + ∠COB = 180°         .....[ linear pair ]
⇒ ∠AOB = ∠ COB= 90° and AC ⊥ BD 

(ii) In ΔAOD and ΔCOD,
OD = OD                ...[ common ]
∠AOD = ∠COD      ...[ each=90° ]
AO = CO                ...[ given]
∴By Side-Angel-Side criterion of congruence, we have
ΔAOD ≅ ΔCOD

(iii) The corresponding parts of the congruent
triangles are congruent.
∴AD = CD             ...[c. p.c t ]
Hence proved. 

Question 11.1

The following figure has shown a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN.
Prove that:  (i) AM = AN  (ii) ΔAMC ≅ ΔANB

Sol:

In ΔABC, AB = AC. m and N are points on
AB and AC such that BM = CN
BN and CM are joined

(i) In ΔAMC and ΔANB
AB = AC               ...[ Given ]  ...(1)
BM = CN              ....[ Given ] ...(2)
Subtracting (2) from (1), we have
AB - BM = AC - CN
⇒ AM = AN                   ...(3)

(ii) Consider the triangles AMC and ANB
AC = AB                      ...[ given ] 
∠AMC =  ∠ANB         ...[ common 90° ]
AM = AN                  ....[ from ( 3 ) ]
∴ By the Side-Angel-Side Criterion of congruence, we have ΔAMC ≅ ΔANB

Question 11.2

The following figure has shown a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN.

Prove that:  (i) BN = CM (ii) ΔBMC≅ΔCNB   

Sol:

In ΔABC, AB = AC. m and N are points on
AB and AC such that BM = CN
BN and CM are joined

(i) The corresponding parts of the congruent triangles are congruent.
∴ CM = BN         ....[ c.p.c.t ] ...(1)

(ii) Consider the triangles ΔBMC and ΔCNB
BM = CN      ...[ given ] 
BC = BC       ...[ common ]
Cm = BN      ..[ from (1) ]
∴ By Side-Side-Side criterion of congruence, we have ΔBMC ≅ ΔCNB 

Question 12

In a triangle, ABC, AB = BC, AD is perpendicular to side BC and CE is perpendicular to side AB.
Prove that: AD = CE.

Sol:

ln ΔABD and ΔCBE,
AB = BC                 ....( given )
∠ ADB = ∠ CEB = 90°  ....[Perpendiculars]

∠B = ∠B                 ....( Common angle )
∴ ΔABD ≅ ΔCBE    ....( by AAS congruence )
⇒ AD = CE            ...( c.p.c.t. )

Question 13

PQRS is a parallelogram. L and M are points on PQ and SR respectively such that PL = MR.
Show that LM and QS bisect each other.

Sol:

Given: PL = RM
To prove: SP = PQ and MP = PL
Proof:
Since SR and PQ are opposite sides of a parallelogram,
PQ = SR                          ...(i)
Also, PL = RM                 ...(ii)
Subtracting (ii) from (i),
PQ - PL = SR - RM
⇒ LQ = SM                     ....(3)
Now, in ΔSMP and ΔQLP,
∠MSP = ∠PQL              ....( alternate interior angles )
∠SMP = ∠PLQ              ....( alternate interior angles )
SM = LQ                       ....[ from(3) ]
∴ ΔSMP ≅ ΔQLP          ....( by ASA congruence )
⇒ SP = PQ and MP = PL     ....( c.p.c.t. )
⇒ LM and QS bisect each other.

Question 14

In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced up to point R so that CR = BP.

Prove that QR bisects PC.
Hint: ( Show that ∆ QBP is equilateral
⇒ BP = PQ, but BP = CR
⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM ).

Sol:

ΔABC is an equilateral triangle,
So, each of its angles equals 60°.
QP is parallel to AC,
⇒ ∠PQB = ∠RAQ = 60°
ln ΔQBP,
∠PQB = ∠BQP = 60°
So, ∠PBQ + ∠BQP + ∠BPQ = 180°   ....(angle sum property)
⇒ 60°+ 60° + ∠BPQ = 180°
⇒ ∠BPQ = 60°
So, ΔBPQ is an equilateral triangle.
⇒ QP = BP 
⇒ QP = CR                                ....(i)
Now, ∠QPM + ∠BPQ = 180°    ...(linear pair)
⇒ ∠QPM+ 60°= 180°
⇒ ∠QPM = 120°
Also, ∠RCM+ ∠ACB = 180°       ...(linear pair)
⇒ ∠RCM+ 60° = 180°
⇒ ∠RCM = 120°
ln ΔRCM and ΔQMP,
∠RCM = ∠QPM                        ....(each is 120°)
∠RMC = ∠QMP                   ...(vertically opposite angles)
QP= CR                                   ....(from(i))
⇒ ΔRCM ≅ ΔQMP  ....(AAS congruence criterion)
So, CM = PM
⇒ QR bisects PC.

Question 15

In the following figure, ∠A = ∠C and AB = BC.
Prove that ΔABD ≅ ΔCBE. 

SOl:

In triangles AOE and COD,
∠A = ∠C                    ...(given)
∠AOE = ∠COD       ...(vertically opposite angles)  
∴ ∠A + ∠AOE = ∠C + ∠COD
⇒ 180° - ∠AEO = 180° - ∠CDO
⇒ ∠AEO = ∠ CDO      ….(i)
Now, ∠AEO + ∠OEB = 180°   ....(linear pair)
And, ∠CDO + ∠ODB = 180°   ....(linear pair)
∴ ∠AEO + ∠OEB = ∠CDO + ∠ODB
⇒ ∠OEB = ∠ODB                     ....[ Using (i) ]
⇒ ∠CEB = ∠ADB                      ….(ii)
Now, in ΔABD and ΔCBE,
∠A = ∠C                                 ....(given)
∠ADB = ∠CEB                         ...[ From (ii) ]
AB = BC                                  ....(given)
⇒ ΔABD ≅ ΔCBE                    ....(by AAS congruence criterion).

Question 16

AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.

Sol:

In ΔAOD and ΔBOC,
∠ AOD = ∠ BOC     ....(vertically opposite angles)
∠ DAO = ∠ CBO      ....(each 90°)
AD = BC                  ....(given)
∴ ΔAOD ≅ ΔBOC    ...(by AAS congruence criterion) 
⇒ AO = BO             ...(c.p.c.t.)
⇒ O is the mid-point of AB.
Hence, CD bisects AB.

Question 17

In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
                   (ii) AO bisects angle BAC.

Sol:

In ΔABC,
AB = AC
⇒ ∠B = ∠C ...( angles opposite to equal sides are equal )

⇒ ${\displaystyle \frac{1}{2}\angle \text{B}=\frac{1}{2}\angle \text{C}}$

⇒ ∠OBC = ∠OCB       ...[ ∵ OB and OC are bisectors of ∠B and ∠C respectively, ∠OBC = ${\displaystyle \frac{1}{2}\angle \text{B}\text{and}\angle \text{OCB}=\frac{1}{2}\angle \text{C}}$ ] ...(i)

⇒ OB = OC              ...( Sides opposite to equal angles are equal )  ...(ii)

Now, in ΔABO and ΔACO,
AB = AC                 ...( given )
∠OBC = ∠OCB       ...[ from(i) ]
OB = OC                ...[ from(ii) ] ...( proved )
∴ ΔABO ≅ ΔACO   ...( by SAS congruence criterion )
⇒ ∠BAO = ∠CAO   ...( c.p.c.t. )
⇒  AO bisects ∠BAC  ...(proved)

Question 18

In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°.

Prove that AD = FC.

SOl:

Given that, BC = DE
⇒ BC + CD = DE + CD  ....( Adding CD on both sides )
⇒ BD = CE                     ....(i)
Now, in ΔABD and ΔFEC,
AB = EF                         ....(given)
∠ABD = ∠FEC               ....(Each 90°)
BD = CE                        ....[ From (i) ]
⇒  ΔABD ≅  ΔFEC         ...(by SAS congruence criterion)
⇒ AD = FC                    ...(c.p.c.t.)

Question 19

A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.

Sol:

In ΔAOD and ΔAOB,
AD = AB                 ...(given)
AO = AO                 ...(Common)
OD = OB                 ...(given) 
⇒ ΔAOD ≅ ΔAOB    ...(by SSS congruence criterion)
⇒ ∠AOD = ∠AOB    ...(c.p.c.t.)  ...(i)
Similarly, ΔDOC ≅ ΔBOC
⇒ ∠DOC = ∠BOC  ...(c.p.c.t.)  ...(ii)

But, ∠AOB + ∠AOD + ∠COD + ∠BOC = 4 Right angles ...[ Sum of the angles at a point is 4 Right angles ]
⇒ 2∠AOD + 2∠COD = 4 Right angles    ....[ Using (i) and (ii) ]
⇒ ∠AOD + ∠COD = 2 Right angles
⇒ ∠AOD + ∠COD = 180°
⇒ ∠AOD and ∠COD form a linear pair.
⇒ AO and OC are in the same straight line.
⇒ AOC is a straight line.

Question 20

In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA

Sol:

Given: In quadrilateral ABCD, AD = BC and BD = AC.

To Prove:

(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA

Proof:

In ΔABD and ΔBAC,
AD = BC           ....(given)
BD = CA           ....(given)
AB = AB           ....(common)

∴ ΔABD ≅ ΔBAC ....(by SSS congruence criterion)

${\displaystyle \begin{array}{l}\angle \text{ADB}=\angle \text{BCA}\\ \angle \text{DAB}=\angle \text{CBA}\end{array}\}...\left(\text{c.p.c.t.}\right)}$