Question

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Answer 1

Step-by-step explanation:

We know that side opposite to greater angle is greater.

So,here in the question,it is given that

AB > AC

i.e.

$\angle \: ACB \: > \angle \: ABC$

Now

$\angle \: ABD = \angle \: ACD+ \angle \: DAC \: \: \\ (external \: angle \: property \: of \: triangle)$

As angle ABC is less than angle ACD

thus

$\angle \: ABD \: > \angle \: ABC \\ \\ thus \\ \\ AB > AD \: \: (side \: opposite \: to \: greater \: angle \: is \: greater)$

Hence proved.

Answer 2

Answer:

AB > AC

∠ACB>∠ABC

now

ABD=∠ACD+∠DAC

As angle ABC is less than angle ACD

As angle ABC is less than angle ACDthus

ABD>ABC

AB>AD

hence proved