Question

if an equilateral triangle ABC,AD is perpendicular on BC then prove that 3AB² = 4AD²

Answer 1

TAKE THAT SIDES
AB = BC = CA = a

then AD CUT BC AS
BD = CD = a/2

SO IN ∆ ADB ANGLE ADB = 90°

AB² = BD²+AD²

(PYTHAGORAS THEOREM)

(a)² = (a/2)²+AD²

( AB = a)

a². = a²/4+AD²

a²-(a²/4) = AD²

4a²-a²/4 = AD²

(LMC of 1 and 4 is 4)

3a²/4 = AD²

(cross multiplication)

3a² = 4AD²

3AB² = 4AD². --------------->hence proved

(AB = a)

Answer 2

Hi...☺

Here is your answer...✌

GIVEN THAT,

ABC is an equilateral triangle

Therefore,

AB = BC = AC

And, AD is prependicular to BC

$\angle ADB = \angle ADC = 90 \degree$

Also,

BD = DC = (1/2) BC

BD = DC = (1/2) AB [ °•° BC = AB ]

Now,

In ∆ADB,
Using Pythagoras Theorem

AB² = AD² + BD²

${AB}^{2} = {AD}^{2} + {( \frac{1}{2}AB )}^{2} \\ \\ AB^{2} - \frac{1}{4} {AB}^{2} = {AD}^{2} \\ \\ \frac{3{AB}^{2}}{4} = {AD}^{2} \\ \\ 3{AB}^{2} = 4{AD}^{2}\: \:[Proved]$