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# 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}

AC^{2 }= AD^{2} + [BC - BD]^{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: **Class 10 Maths NCERT Solutions Chapter 6

**Video Solution:**

## In Fig. 6.59, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that: AC² = AB² + BC² - 2BC.BD.

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.6 Question 4

**Summary:**

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.

**☛ Related Questions:**

- 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.
- In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) ΔPAC ~ ΔPDB (ii) PA.PB = PC.PD.

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