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
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°
AC2 = AD2 + CD2 [Using Pythagoras theorem]
AD2 = AC2 - CD2.....(3)
In ∆ADB, ∠ADB = 90°
AB2 = AD2 + BD2 [Using Pythagoras theorem ]
AB2 = AC2 - CD2 + BD2 [from equation(3)]
AB2 = AC2 + (3/4 BC)2 - (1/4 BC)2 [from equations(1) and (2)]
AB2 = AC2 + (9BC2 - BC2)/16
AB2 = AC2 + 8BC2/16
AB2 = AC2 + 1/2 BC2
Thus, 2AB2 = 2AC2 + BC2
☛ 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 2AB2 = 2AC2 + BC2.
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