# 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:**

ΔADC is a right triangle as ∠ADC = 90°

Using Pythagoras theorem,

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

AC^{2 }= AD^{2} + (BD + BC)^{2}

AC^{2 }= AD^{2} + BD^{2} + BC^{2} + 2BC × BD

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

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

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## In Fig. 6.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC.BD

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.6 Question 3

**Summary:**

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

**☛ Related Questions:**

- In Fig. 6.59, ABC is a triangle in which ∠ABC less than 90° and AD ⊥ BC. Prove that: AC² = AB² + BC² - 2BC × BD.
- In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that : (i) AC² = AD² + BC.DM + (BC/2)² (ii) AB² = AD² - BC.DM + (BC/2)² (iii) AC² + AB² = 2AD² + 1/2BC²
- Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
- In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that: (i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP.