In △ ABC it is given that ∠ A = 70o and ∠ B = 60o
Based on the sum property of the triangle
∠ A + ∠ B + ∠ C = 180o
To find ∠ C
∠ C = 180o – ∠ A – ∠ B
By substituting the values in the above equation
∠ C = 180o – 70o – 60o
∠ C = 180o – 130o
By subtraction
∠ C = 50o
Consider △ BCD
We know that ∠ CBD is the exterior angle of ∠ ABC
So we get
∠ CBD = ∠ DAC + ∠ ACB
By substituting the values in the above equation
∠ CBD = 70o + 50o
By addition
∠ CBD = 120o
It is given that BC = BD
So we can write it as
∠ BCD = ∠ BDC
Based on the sum property of the triangle
∠ BCD + ∠ BDC + ∠ CBD = 180o
So we get
∠ BCD + ∠ BDC = 180o – ∠ CBD
By substituting values in the above equation
∠ BCD + ∠ BDC = 180o – 120o
∠ BCD + ∠ BDC = 60o
It can be written as
2 ∠ BCD = 60o
By division
∠ BCD = ∠ BDC = 30o
In △ ACD
It is given that ∠ A = 70o and ∠ B = 60o
We can write it as
∠ ACD = ∠ ACB + ∠ BCD
By substituting the values we get
∠ ACD = 50o + 30o
By addition
∠ ACD = 80o
So we get to know that ∠ ACD is the greatest angle and the side opposite to it i.e. AD is the longest side.
Therefore, it is proved that AD > CD
We know that ∠ BDC is the smallest angle and the side opposite to it i.e. AC is the shortest side.
Therefore, it is proved that AD > AC.