gauravategmat wrote:

@bunnel

I think statement 1 alone is able to prove the same.

In traingle ACB, applying the exterior angle property angle ACB = angle ADC + angle CAD

Statement 1 : ACB = 2.ADC

if ADC = x -> ACB = 2x and thus CAD = x

Now BAD = 180-(90+x) = 90-x

Also, in Traingle ABC , angle BAC = 180-(90+2x) = 90-2x

So, angle BAC = BAD-CAD = 90-x - 90 + 2x = x

so BAC = x and BCA = 2x

so sufficient to prove ABC is not isoseles

Please correct me if I am wrong in any step

Hi

you have written that:

angle BAD = 90-x

and in triangle ABC, angle BAC = 90-2x . I completely agree till here. But after that you have written:

angle BAC = angle BAD - angle CAD = ... here you have taken CAD as (90-2x).. I don't understand how?

CAD=x. Its NOT equal to 90-2x.

I think this step is incorrect. Or if I am missing something, please let me know. Thanks