# In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)

**Solution:**

As we all know the Basic Proportionality Theorem (B.P.T) or (Thales Theorem)

Two triangles are similar if

- Their corresponding angles are equal
- Their corresponding sides are in the same ratio (or proportion)

In, ΔABC

BC || DE

In ΔABC & ΔADE

∠ABC = ∠ADE [corresponding angles]

∠ACB = ∠AED [ corresponding angles]

∠A = ∠A common

⇒ ΔABC ~ ΔADE

AD/DB =-AE/EC

1.5/3 = 1/EC

EC =-3 × 1/1.5

EC = 2cm

(ii) Similarly,

ΔABC ∼ ΔADE

AD/DB = AE/EC

AD/7.2 = 1.8/5.4

AD = (7.2 × 1.8)/5.4

AD = 2.4cm

**Video Solution:**

## In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)

### Class 10 Maths NCERT Solutions - Chapter 6 Exercise 6.2 Question 1:

In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)

For the above figure the missing sides are EC = 2cm and AD = 2.4 cm