Question

IN AN EQUILATERAL TRIANGLE ABC AD PERPENDICULAR TO BC PROVE 3AB2 =4AD2

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Given:

∆ABC is an equilateral triangle.

AD perpendicular to BC.

To Prove:

3AB² = 4AD²

Proof:

In ∆ADB and ∆ADC,

<ADB = <ADC =90° (R)

AB = AC ( hypotenuse)(H)

AD = AD ( common side )(S)

Therefore,

∆ADB congruent to ∆ADC.(RHS criterion )

BD = DC [ CPCT ]

BD = AB/2 --(1)

Now,

In ∆ADB,

AB² = AD² + BD²

$\implies AB^{2}=AD^{2}+\left(\frac{AB}{2}\right)^{2}$

$\implies AB^{2}=AD^{2}+\frac{AB^{2}}{4}$

$\implies AB^{2}-\frac{AB^{2}}{4}=AD^{2}$

$\implies \frac{4AB^{2}-AB^{2}}{4}=AD^{2}$

$\implies 3AB^{2}=4AD^{2}$

Hence , Proved.

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