ΔABC is an equilateral triangle and point M is the midpoint of the altitude AD.
Show that 16BM 2 = 7BC2.
Answers
Answer:
Let AB,BC,ACBE equal
to a
D is the midpoint of
BC.
By Pythagoras theorem
AB
2
=AD
2
+BD
2
a
2
=AD
2
+(a/2)
2
a
2
−
4
a
2
=AD
2
4
3a
2
=AD
2
But AB =a
3AB
2
=4AD
2
∴ 4AD
2
=3AB
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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