Given: In ∆ABC, BD=BC and ∠B=60° and ∠A=70°

To prove: AD>CD and AD>AC

Proof:

In ∆ABC, by the angle sum property, we have

∠A + ∠B + ∠C = 180°

70° + 60° + ∠C = 180°

130° + ∠C = 180°

∠C = 50°

Now in ∆BCD we have,

∠CBD = ∠DAC + ∠ACB … as ∠CBD is the exterior angle of ∠ABC

= 70° + 50°

Since BC = BD …given

So, ∠BCD = ∠BDC

∴ ∠BCD + ∠BDC = 180° - ∠CBD

= 180° - 120° = 60°

2∠BCD = 60°

∠BCD = ∠BDC = 30°

Now in ∆ACD we have

∠A = 70°, ∠D = 30°

And ∠ACD = ∠ACB + ∠BCD

= 50° + 30° = 80°

∴∠ACD is greatest angle

So, the side opposite to largest angle is longest, ie AD is longest side.

∴AD > CD

Since, ∠BDC is smallest angle,

The side opposite to ∠BDC, ie AC, is the shortest side in ∆ACD.

∴AD > AC