Question

In fig.6, D is a point on side BC of ∆ABC such that AD = AC. Show that AB >
AD.

Answer 1

@vishweshshrimali5 @hartnn @Callisto

Answer 2

I am just stuck in this question ...

Answer 3

Prove using contradiction method

Answer 4

Let AD = AB then

Answer 5

in triangle ABD and triangle AD = AC

Answer 6

Ok

Answer 7

Well I did

Answer 8

AB = AD = AC
and
AD is common
BD = DC
Then by SSS congruency

Answer 9

both triangles are congruent

Answer 10

BD= DC ..? any prove

Answer 11

Or u can prove angleADB>angleABD

Answer 12

sorry

Answer 13

AB = AD = DC

Answer 14

or :
since D lies between B and C
we can write 2 things :
BC>DC ----->(1)
perimeter of ABC > ADC --->(2)
from (2)
AB+BC+AC>AD+DC+AC
implies
AB+BC>AD+DC
from (1) since BC>DC
AB> AD

Answer 15

is it given that AD = DC ? i think you meant AD = AC

Answer 16

was that possible? that if AB +BC > AD + DC
then we can say that AB > AD

oh yes...

Answer 17

means angle ADC = angle ACD
angle ABD = angle ADB

also
angle ADB + angle ADC = 180
thus
angle ABD + angle ACD = 180

which is not possible

Answer 18

just because BC>DC

Answer 19

hmn right

Answer 20

I got it now @vishweshshrimali5 and @hartnn

Answer 21

Can any one here give me a favor to give a medal to hartnn

Answer 22

there is a contradiction because sum of 3 angles of triangle is 180 whereas here are only 2 angles' sum 180
So AD < AB ( becuase AD can't be greater than AB as that will mean that point D is outside the triangle):(don't forget to write this line)

Answer 23

done :D

Answer 24

thanks ..

Answer 25

;)

Answer 26

He deserved it :)

Answer 27

personally i am not comfortable with contradiction when we can prove without it,but vushwesh gave a nice proof with it.

Answer 28

Yes I agree with hartnn

Contradiction is generally used when we are sure that we can prove the opposite wrong..