In a triangle ABC, DE is parallel to BC; AD = a, DB = a + 4, AE = 2a + 3, EC = 7a. What is the value of 'a’ if a>0 ?

$$\triangle ADE and \triangle ABC are similar triangle.

($$\because all \angle A is common and \angle B = \angle D & \angle C = \angle E$$)

By the properties,

$$\frac{AB}{AD} = \frac{AC}{AE}$$

$$\frac{AB}{a} = \frac{AC}{2a + 3}$$

$$\frac{AD + DB}{a} = \frac{AE + EC}{2a + 3}$$

$$\frac{a + a + 4}{a} = \frac{2a + 3 + 7a}{2a + 3}$$

$$\frac{2a + 4}{a} = \frac{9a + 3}{2a + 3}$$

{2a + 4}{2a + 3} = {a}{9a + 3}

$$4a^2 + 14a + 12 = 9a^2 + 3a$$

$$5a^2 - 11a - 12 = 0$$

$$5a^2 - 15a + 4a - 12 = 0$$

$$5a(a - 3)+ 4(a - 3) = 0$$

(5a + 4)(a - 3) = 0

a = -4/5 and a = 3

$$\because$$ a>0 so, a = 3