In ` A B C ,` side `A B` is produced to `D` so that `B D=b cdot` If `/_B=60^0` and `/_A=70^0,` prove that : `(i) A D > C D` `(ii)A D > A C`

Explanation

Given that in `triangleABC`, side `AB` is produced to `D` So that `BD = BC` and `angleB = 60^0` and `angleA = 70^0`,
We have to prove that
(i) `AD > CD`
(ii)` AD > AC`
First join `C` and `D`
Now, in `triangleABC`
Sum of angles in a triangle=`180^0`
`angleAangleB+angleC=180^0`
`angleC=180^0-70^0-60^0`
`180^0-130^0`
`angleC=50^0`
`angleACB=50^0`.........(i)
And also in `triangleBDC`,
`angleDBC=180^0-angleABC`............(ABC is a staight angle)
`180^0-60^0`
`120^0`
And also `BD=BC`.......(given)
`angleBCD=angleBDC`
Angle oopposite to equal sides are equal
Now,
Sum of angle in a triangle =`18^0`
`angle DBC+ange BCD+angleBDC=180^0`
`120^0+angleBCD+angleBCD=180^0`
`2angleBCD=180^0-120^0`
`angleBCD=60/2=30^0`
`angleBCD=angleBDC=30^0`............(ii)
Now, consider `triangleADC`
`angleBAC=angleDAC=70^0`........(given)
`angleBDC=angleADC=30^0`......(from (i) and (ii))
`angleACD=angleACB+angleBCD`
`50^0+30^0`......from equation (i) and (ii)
`80^0`
Now ,`angleADC < angleDAC < angleACD`
`AC`AD>CD` and `AD>AC`
Hence proved