# The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB^{2} = 2AC^{2} + BC^{2}

**Solution:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Theorem 6.8

In ∆ABC , AD ⊥ BC and BD = 3CD

BD + CD = BC

3CD + CD = BC

4CD = BC

CD = 1/4 BC .....(1)

and, BD = 3/4 BC.....(2)

In ∆ADC

AC^{2} = AD^{2} + CD^{2} [∵ ∠ADC = 90°]

AD^{2} = AC^{2} - CD^{2}.....(3)

In ∆ADB

AB^{2} = AD^{2} + BD^{2} [∵ ∠ADB = 90°]

AB^{2} = AC^{2} - CD^{2} + BD^{2} [from (3)]

AB^{2} = AC^{2} + (3/4BC)^{2} - (1/4 BC)^{2} [from (1) and (2)]

AB^{2} = AC^{2} + (9BC^{2} - BC^{2})/16

AB^{2} = AC^{2} + 8BC^{2}/16

AB^{2} = AC^{2} + 1/2 BC^{2}

2AB^{2} = 2AC^{2} + BC^{2}.

**Video Solution:**

## The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB^{2} = 2AC^{2} + BC^{2}

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.5 Question 14:

The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see Fig. 6.55). Prove that 2AB^{2} = 2AC^{2} + BC^{2}

The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD in the above figure. It is proved that 2AB^{2} = 2AC^{2} + BC^{2}