Question

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

Answer 1

In ∆ABC

AB > AC

⇒ ∠ACB > ∠ABC …(i) (∵ angle opposite to larger side is greater) .

Now in ∆ACD, CD is produced to B forming an exterior ∠ADB.

∠ADB > ∠ACD (exterior angle of a A is greater than each interior opposite angle)

⇒ ∠ADB > ∠ACB …(ii)

∠ACD = ∠ACB

From (i) and (ii)

∠ADB > ∠ABC

⇒ ∠ADB > ∠ABD [∵ ∠ABC = ∠ABD]

⇒ AB > AD [∵ side opposite to greater angle is larger]

Hence proved

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