Question

In an equilateral triangle ABC, AM perpendicular to BC.
Prove that : 3AB^2 = 4AM^2​

Answer 1

Given:

A ∆ ABC, in which sides are AB=BC= AC= a units

and AD is perpendicular to BC ,

In ∆ADB ,

AB²= AD²+ BD²     (by Pythagoras theorem)

a² = AD² + (a/2)²   [BD= 1/2BC, since in an equilateral triangle altitude AD is  perpendicular bisector of BC ]

a²- a²/4 =AD²

=( 4a²-a²)/4 = AD²

= 3a² /4 = AD²

3AB²/4= AD²

[ AB= a]

3AB²= 4AD²

[FIGURE IS IN THE ATTACHMENT]

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Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

∆ABC is an equilateral triangle.

AM is the altitude of the triangle.

TO PROVE :

3AB^2 = 4AM^2

PROOF :

We know in an equilateral triangle altitude divides the base in two equal parts.

$\therefore$ BM = CM = 1/2 BC

Now, In rt. ∆AMB

$\implies \: \sf{{AB}^{2} = {AM}^{2} + {BM}^{2}}$

$\implies \: \sf{{AB}^{2} = {AM}^{2} + { (\frac{BC}{2} )}^{2}}$

$\implies \: \sf{{AB}^{2} = {AM}^{2} + { (\frac{AB}{2}) }^{2} } \: \: \: \: \: [AB = BC]$

$\implies \: \sf{{AB}^{2} = {AM}^{2} + { \frac{AB}{4} }^{2} }$

$\implies \: \sf{{4AB}^{2} = {4AM}^{2} + {AB}^{2}}$

$\implies \: \sf{{3AB}^{2} = {4AM}^{2} }$

Hence Proved.