PROBLEMS INVOLVING ANGLE BISECTOR THEOREM

BD bisects <ABC. Find the value of x :

Example 1 :

Solution :

In the given figure, BD bisects <ABC

So, <ABD  =  <CBD

(x + 15)˚  =  (4x – 45)˚

x + 15˚  =  4x - 45˚

x – 4x  =  -45 – 15

-3x  =  -60

x  =  20˚

So, the value of x is 20˚

Example 2 :

Solution :

In the given figure, BD bisects <ABC

So, <ABD  =  <CBD 

(2x + 35)˚  =  (5x – 22)˚

2x + 35˚  =  5x - 22˚

2x – 5x  =  -35 – 22

-3x  =  -57

x  =  19˚

So, the value of x is 19˚

Example 3 :

Solution :

In the given figure, BD bisects <ABC

So, <ABD  =  <CBD

(10x - 51)˚  =  (6x – 11)˚

10x - 51˚  =  6x - 11˚

10x – 6x  =  51˚ – 11˚

4x  =  40˚

x  =  10˚

So, the value of x is 10˚

Example 4 :

Solution :

In the given figure, BD bisects <ABC

So, <ABD  =  <CBD

(2x + 7)˚  =  (4x – 9)˚

2x + 7˚  =  4x - 9˚

2x – 4x  =  -7 – 9

-2x  =  -16

x  =  8˚

So, the value of x is 8˚

Example 5 :

Solution :

In the given figure, BD bisects <ABC

So, <ABD  =  <CBD

(15x + 18)˚  =  (23x – 14)˚

15x + 18˚  =  23x - 14˚

15x – 23x  =  -18 – 14

-8x  =  -32

x  =  4˚

So, the value of x is 4˚

Example 6 :

Solution :

In the given figure, BD bisects <ABC

So, <ABD  =  <CBD

(1/2x + 20)˚  =  (3x – 85)˚

1/2x + 20˚  =  3x - 85˚

1/2x – 3x  =  -20 – 85

-5/2x  =  -105

5x  =  210

x  =  42˚

So, the value of x is 42˚

Example 7 :

The bisector of <A of the triangle ABC meets BC at D and BC is produced to E. Prove that <ABC + <ACE = 2 <ADC

Solution :

Since AD is the bisector of <BAC,

<BAD = <DAC = x

<ACE = exterior angle 

= sum of remote interior angles

Let <ABC = y ------(1)

<ABC + <BAC = <ACE

y + 2x = <ACE

<ACE = 2x + y ------(2)

(1) + (2)

<ABC + <ACE = y + 2x + y

= 2y + 2x

= 2(x + y)

= 2 (<BAD + <ABC)

Example 8 :

The line segment joining the midpoint of any side of with opposite vertex is

a) altitude     b)  median      c) Perpendicular bisector

d) angle bisector

Solution :

Midpoint of line segment will be at the middle exactly. The line segment joining the midpoint of any side of the opposite vertex is at median.

Example 9 :

Triangles ABC and DBC are two isosceles triangles on the same base BC and their vertices A and D are also on the same side of BC. If AD is extended to intersect BC at P, show that

(i) ∆ABD ≅ ∆ACD

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects angles BAC and BDC

(iv) AP is the perpendicular bisector of BC

Solution :

(i) 

Triangle ABD and ACD are isosceles triangles. Then AB and AC are congruent sides.

AB = AC

AD is common

Since BDC is an isosceles triangle, BD = DC

Then, ∆ABD ≅ ∆ACD

(ii)

In triangle ABP and ACP,

AB = AC

AP is common

<ABP = <ACP

Then, ∆ABP ≅ ∆ACP

(iii) Using CPCTC, from the above proof

<BAP = <CAP

<BDP = <CDP

Then AP bisects angles BAC and BDC.

(iv)

<BPD + <CPD = 180

Since DBP and DCP are congruent, then 

<BPD = <CPD (by CPCTC)

<BPD + <BPD = 180

2<BPD = 180

<BPD = 90

Then AP is the perpendicular bisector of BC.

Example 10 :

In quadrilateral ABCD, AC = AD and AB bisects <A and triangle ABC

a)  BC > BD       b)  BC < BD

c)  BC = BD      d) BC = (1/2 )BD

Solution :

In triangles ACB and ADB

<CAB = <DAB (AB bisects <CAD)

AC = AD (Given)

AB = AB (Common)

Then, triangles ACB and ADB are congruent. Here BC and BD are corresponding sides of congruent triangles. So, BC = BD.

Example 11 :

In the given figure AD is perpendicular to BC. AE is the angle bisector of <BAC. Find <DAE.

Solution :

In triangles ABD,

<BDA = 90, <DBA = 60 and <BAD = 30

In triangle ADC, 

<ADC + <DCA + <DAC = 180

90 + 35 + <DAC = 180

<DAC = 180 - 125

<DAC = 55

<DAE = <DAC / 2

= 55/2

= 27.5

Example 12 :

EF is the bisector of DEG(not drawn in scale). If <DEF = 3x + 1 and <DEG = 5x + 19 , find the value of x.

Solution :

<DEF = 3x + 1 and <DEG = 5x + 19

<DEG / 2 = <DEF

(5x + 19)/2 = 3x + 1

5x + 19 = 2(3x + 1)

5x + 19 = 6x + 3

19 - 3 = 6x - 5x

16 = x

So, the value of x is 16.

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