In a △ABC, ∠A:∠B:∠C=2:3:4.A line drawn parallel to AB , then the ∠ACD is.

Correct option is A

We are tasked to determine $$\angle$$​ ACD in $$\triangle$$​ ABC when a line is drawn parallel to AB and the ratio of angles $$\angle A : \angle B : \angle C$$​ = 2 : 3 : 4 .

Step-by-Step Solution:

1. Given:

- Ratio of angles $$\angle A : \angle B : \angle C = 2 : 3 : 4$$​

- Line parallel to AB is drawn.

2. Calculate the angles of \triangle ABC :

- The sum of angles in a triangle is $$180^\circ$$​ .

- Let k be the common multiplier. Then:

$$\angle A = 2k, \quad \angle B = 3k, \quad \angle C = 4k$$​

2k + 3k + 4k = $$180^\circ$$​

$$9k = 180^\circ \quad \Rightarrow \quad k = 20^\circ$$​

- Substituting k :

$$\angle A = 40^\circ, \quad \angle B = 60^\circ, \quad \angle C = 80^\circ$$​

3. Properties of the parallel line:

- The line drawn parallel to AB forms alternate interior angles with $$\triangle$$​ ABC .

- Hence, $$\angle ACD = \angle A$$​

Final Answer:

$$\boxed{\angle ACD = 40^\circ}$$​