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# 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.

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, ∠ADC = 90°

AC^{2} = AD^{2} + CD^{2} [Using Pythagoras theorem]

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

In ∆ADB, ∠ADB = 90°

AB^{2} = AD^{2} + BD^{2} [Using Pythagoras theorem ]

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

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

Thus, 2AB^{2} = 2AC^{2} + BC^{2}

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

**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 2AB² = 2AC² + BC²

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 14

**Summary:**

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

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