Question

pls help me to slove this​

Answer 1

∠ADC >  ∠ADB  if AC > AB  and AD is bisector of ∠A

Step-by-step explanation:

given that AC > AB

in Δ ABC

AC > AB

=> ∠B > ∠C   as greater sides has opposite greater angles

AD bisect ∠A

=> ∠BAD = ∠CAD  = ∠A/1

in Δ ABD

∠B + ∠ADB + ∠BAD = 180°

in Δ ACD

∠C + ∠ADC + ∠CAD = 180°

Equating

∠B + ∠ADB + ∠BAD = ∠C + ∠ADC + ∠CAD

as ∠BAD = ∠CAD

=> ∠B + ∠ADB = ∠C + ∠ADC

=>  ∠B  - ∠C = ∠ADC -  ∠ADB

as  ∠B > ∠C => ∠ADC >  ∠ADB

Hence Proved

QED

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