Question

In triangle ABC, AB > AC and D is a point in side BC. Show that : AB > AD.

Answer 1

Answer:

In ΔDAC,

AD=AC (Given)

So, ∠ADC=∠ACD (Angles opposite to equal sides)

Now, ∠ADC is an exterior angle for △ABD.

So, ∠ADC>∠ABD

∠ACD>∠ABD

∠ACB>∠ABC

AB>AC (Side opposite to larger angle in △ABC)

AB>AD

Step-by-step explanation:

Hope it helps you

Answer 2

Step-by-step explanation:

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∠ADB=∠DAC+∠C

⇒∠ADB>∠C

⇒∠ADB>∠C>∠B

⇒∠ADB>∠B

⇒AB>AD

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