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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
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 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 perpendicular drawn from vertex to opposite side bisects the side
Thus, BE = CE = 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
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
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
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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