Question

In the figure D is any point on the side BC of triangle ABC, If AB> AC show that AB > AD.​

Answer 1

Answer:

angle ADB > angle ABD

so side opposite to larger angle is greater

so AB>AD

Step-by-step explanation:

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Answer 2

$\huge\color{blue}\boxed{\colorbox{white}{★Solution★}}$

°°° Explanation °°°

In ∆ABC AB > AC ⇒ ∠ACB > ∠ABC …

• (i) (∵ angle opposite to larger side is greater) Now in ∆ACD, CD is produced to B forming an ext ∠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. ]