# In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that: AC^{2} = AB^{2} + BC^{2} + 2BC.BD

**Solution:**

Using Pythagoras theorem in In ΔADC as ∠ADC = 90°

AC^{2} = AD^{2} + CD^{2}

= AD^{2} + [BD + BC]^{2}

= AD^{2} + BD^{2} + BC^{2} + 2BC × BD

= AB^{2} + BC^{2} + 2BC × BD [∵ In DADB, AB^{2} = AD^{2} + BD^{2}]

Hence it is proved that: AC^{2} = AB^{2} + BC^{2} + 2BC.BD.

**Video Solution:**

## In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that: AC^{2} = AB^{2} + BC^{2} + 2BC.BD

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.6 Question 3:

In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that: AC^{2} = AB^{2} + BC^{2} + 2BC.BD

In the above figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Hence it is proved that: AC^{2} = AB^{2} + BC^{2} + 2BC.BD