Question

TriangleABC is an equilateral triangle and point M is the midpoint of the altitude AD.
Show that 16 BM2 = 7BC2

Answer 1

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Answer 2

Given :-

• ∆ABC is a equaliteral ∆ .
• M is the mid point of altitude AD .

To Show :-

• 16BM² = 7BC² .

Solution :-

since , AD is altitude at BC .

so,

→ ∠ADB = 90° .

then, in right ∆ADB we have,

→ AD² + DB² = AB² ---------- Eqn.(1) { By pythagoras theorem.}

similarly, in right ∆MDB we have,

→ MD² + DB² = BM² ---------- Eqn.(2)

now, given that, M is mid point of Altitude AD .

so,

• AM = MD = (1/2)AD

putting value of MD in terms of AD in Eqn.(2) , we get,

→ {(1/2)AD}² + DB² = BM²

→ (1/4)AD² DB² = BM²

→ AD² + 4•DB² = 4•BM²

→ (AD² + DB²) + 3•DB² = 4•BM²

putting value from Eqn.(1),

→ AB² + 3•DB² = 4•BM²

→ AB² = 4•BM² - 3•DB²

now, since , ∆ABC, is an Equaliteral ∆ . we have,

• AB = BC = CA
• Altitude of an Equaliteral ∆ bisects the base . so, BD = (1/2)BC .

using both we get,

→ BC² = 4•BM² - 3{(1/2)BC}²

→ BC² = 4•BM² - (3/4)BC²

→ BC² = (16•BM² - 3•BC²)/4

→ 4•BC² = 16•BM² - 3•BC²

→ 4•BC² + 3•BC² = 16•BM²

→ 7BC² = 16BM²

→ 16BM² = 7BC² (Proved).

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