Question

In a triangle ABC, AC > AB and bisector of angle A meets BC
at D then angle ADB is:​

Answer 1

Correct question :- In a triangle ABC, AC > AB and bisector of angle A meets BC at D then show that ADC is greater than ADB ?

Solution :-

given that,

• AC > AB.
• AD is bisector of ∠A

So,

→ ∠BAD = ∠CAD

Also in ∆ABD by angle sum Property, we have,

→ ∠ADB = 180° - {∠ABD + (∠A/2)}

Now, we know that, angle opposite to longer side is greater .

therefore,

→ ∠ABD > ∠ACD

→ ∠ABD + (∠A/2) > ∠ ACD + (∠A/2)

→ 180° - {∠ACD + (∠A/2)} > 180° - {∠ABD +(∠A/2)}

→ ∠ADC > ∠ADB . (Proved.)

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Answer 2

Answer:

it is acute angle as other angle is obtuse