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

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