# In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that: AC^{2} = AB^{2} + BC^{2} - 2BC × BD

**Solution:**

Using Pythagoras Theorem In ΔADC as ∠ADC = 90°

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

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

= AD^{2} + BD^{2} + BC^{2} - 2BC.BD

AC^{2} = 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. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that: AC^{2} = AB^{2} + BC^{2} - 2BC × BD

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.6 Question 4:

In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that: AC^{2} = AB^{2} + BC^{2} - 2BC × BD

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