Given: ∠BAD = ∠DAC

To prove: AB>BD and AC>DC

Proof:

In ∆ACD,

∠ADB = ∠DAC + ∠ACD … exterior angle theorem

= ∠BAD + ∠ACD … given that ∠BAD = ∠DAC

∠ADB > ∠BAD

The side opposite to angle ∠ADB is the longest side in ∆ADB

So, AB > BD

Similarly in ∆ABD

∠ADC = ∠ABD + ∠BAD … exterior angle theorem

= ∠ABD + ∠CAD … given that ∠BAD = ∠DAC

∠ADC > ∠CAD

The side opposite to angle ∠ADC is the longest side in ∆ACD

So, AC > DC