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In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD2 = 7AB2

Solution:

We know that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD2 = 7AB2.

In ΔABC as shown in the figure above,

AB = BC = CA and BD = 1/3 BC

Draw AE ⊥ BC

We know that in an equilateral triangle, a perpendicular drawn from the vertex to the opposite side bisects the side.

Thus, BE = EC = 1/2 BC

Now, in ΔADE,

AD2 = AE2 + DE2 (Pythagoras theorem) ---------------- (1)

AE is the height of an equilateral triangle which is equal to √3/2 side

Thus, AE = √3/2 BC

Also, DE = BE - BD [From the diagram]

Substituting these in equation(1) we get,

AD2 = (√3/2 BC)2 + (BE - BD)2

AD2 = 3/4 BC2 + [BC/2 - BC/3]2 ---------------- (BE can be written as BC/2 and BD can be written as BC/3)

AD2 = 3/4 BC2 + (BC/6)2

AD2 = 3/4 BC2 + BC2/36

AD2 = (27BC2 + BC2) / 36

36AD2 = 28BC2

9AD2 = 7BC2

9AD2 = 7AB2 [Since AB = BC = CA]

☛ Check: NCERT Solutions for Class 10 Maths Chapter 6

Video Solution:

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9AD² = 7AB².

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 15

Summary:

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Hence proved that 9AD2 = 7AB2.

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