AD is the bisector of angle A of trianleABC, where D lies on BC.prove tht AB>BD and AC>CD.

Given: In ΔABC

AD is the bisector of ∠A

⇒ ∠BAD = ∠CAD  ..... (1)

In ΔADC

Ext.  ∠ADB > ∠DAC  (exterior angle theorem)

⇒ ∠ADB > ∠BAD  (from (1) )

⇒ AB > BD  (side opposite to greater angle is longer)

Similarly in ΔADB

Ext ∠ADC > ∠DAB

⇒ ∠ADC > ∠DAC

⇒ AC > CD