Given that in , side AB is produced to D so that BD = BC and ∠B = 60^o, ∠A = 70^o

We have to prove that,

(i) AD > CD

And, (ii) AD > AC

First join C and D

Now,

In ,

∠A + ∠B + ∠C = 180^o (Sum of all angles of triangle)

∠C = 180^o – 70^o – 60^o

= 50^o

∠C = 50^o

∠ACB = 50^o (i)

And also in ,

∠DBC = 180^o - ∠ABC (Therefore, ∠ABD is straight angle)

= 180^o – 60^o

= 120^o

BD = BC (Given)

∠BCD = ∠BDC (Therefore, angle opposite to equal sides are equal)

Now,

∠DBC + ∠BCD + ∠BDC = 180^o (Sum of all sides of triangle)

120^o + ∠BCD + ∠BCD = 180^o

2∠BCD = 180^o – 120^o

2∠BCD = 60^o

∠BCD = 30^o

Therefore, ∠BCD = ∠BDC = 30^o (ii)

Now, consider ,

∠BAC = ∠DAC = 70^o (Given)

∠BDC = ∠ADC = 30^o [From (ii)]

∠ACD = ∠ACB + ∠BCD

= 50^o + 30^o [From (i) and (ii)]

= 80^o

Now,

∠ADC < ∠DAC < ∠ACD

AC < DC < AD (Therefore, side opposite to greater angle is longer and smaller angle is smaller)

AD > CD

And,

AD > AC

Hence, proved

We have,

∠ACD > ∠DAC

And,

∠ACD > ∠ADC

AD > DC

And,

AD > AC (Therefore, side opposite to greater angle is longer and smaller angle is smaller)