Question

(b) In the figure (ii) given below, D is any point on the side BC of triangle ABC. If
AB > AC, show that AB > AD.
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Answer 1

Answer:

AD=AC ( Given )

So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]

Now ∠ACD is exterior angle of △ABD

∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]

So, ∠ADC>∠ABD

∠ACD>∠ABD ( from (1))

AB>AC [ side opposite to greater angle is longer ]

∴AB>AD ( As AC=AD given )

Hence, proved.

Answer 2

Answer:

hope it helps thanks it is easy no it's hard