Question

$\huge\underline\mathtt{♡Answer it♡}$​

Answer 1

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

∆ ABC is an equilateral triangle. D is a point on AC. BC is joined.

It is required to prove that, BC > BD.

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

In the ∆ ABD, ∠ BDC is the exterior angle,

∴ ∠BDC > ∠BAC

But since BC = AC

∴ ∠BAC = ∠BCA

∴ ∠BDC > ∠BCA

i. e, ∠BDC > ∠BCD

Hence, BC > BD.

$\huge\underline\mathtt{♡Be Brainly♡}$

Answer 2

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

refer to the attachment