In ?ABC AC is greater than AB angle A bisects meet D prove angle ADc is greater than angle ADB
Dear student,
Given : In ΔABC
AC > AB
and AD is the bisector of ∠A
⇒ ∠BAD = ∠CAD =
Also In ΔABD
∠ABD + ∠BAD + ∠ADB = 180°
⇒ ∠ADB = 180° –
In ΔADC
∠ACD + ∠CAD + ∠ADC = 180°
⇒ ∠ADC = 180° –
Now
∠ABD > ∠ACD (angle opposite to longer side is greater)
Regards