Question

in figure 12.1 11 if AD is the bisector of angle A show that ab greater than BD AC greater than CD ​

Answer 1

Step-by-step explanation:

Given- AD bisects angle A

so, angle 1 = angle 2

we use exterior angle property of a triangle

in this property exterior angle is greater than the interior angle

In ∆ ABD

angle 3 = angle 2 + angle C. { exterior ang. prop.}

we use exterior angle property of a triangle angle 3 is the exterior angle and angle 2 is interior angle

so, angle 3 is greater than the angle 2

angle 2 = angle 1 { given }

so we can put angle 2 to angle 1

so,. angle 3 is greater than the angle 1

the opposite side of angle 3 = AB

and

the opposite side of angle 1 = BD

we can put the opposite sides of angle 3 and angle 1

so, angle 3 is greater than the angle 1

AB is greater than the BD { Hence proved }

similarly ,

we use exterior angle property of a triangle

so, angle ADC = angle 1 + angle B

we use exterior angle property of a triangle angle ADC is the exterior angle and angle 1 is interior angle

so, angle ADC is greater than the angle 1

angle 2 = angle 1

so we can put angle 1 to angle 2

so,. angle ADC is greater than the angle 2

the opposite side of angle ADC = AC

and

the opposite side of angle 2 = CD

we can put the opposite sides of angle ADC and angle 2

so, angle ADC is greater than the angle 2

AC is greater than the CD { Hence proved }