Question

$\huge\underline\mathtt{Answer it}$​

Answer 1

➠ $\huge\underline\mathtt{solution:-}$

D is any point on the extended side BC of the equilateral triangle ABC . AD is joined.

it is required to prove that ∠ BAD > ∠ADB.

$\huge\underline\mathtt{Proof:-}$

ABC is equilateral triangle .

∴ BC = AB

BD = BC + CD

∴ BD > AB (*BC=AB)

∴ In ∆ABD, BD >AB

∴ ∠BAD >∠ADB ( Proved)...

$\huge\underline\mathtt\color{gold}{Be Brainly}$

Answer 2

$\huge\mathtt\blue{Answer:-}$

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