Question

In an equilateral triangle ABC, D is a point on side BC such that BD =1/3BC
BC. Find the value of AD2​

Answer 1

$\huge \mathcal \red {\underline{\underline{A}}} \huge \mathcal \green {\underline{\underline{N}}} \huge \mathcal \pink {\underline{\underline{S}}} \huge \mathcal \blue {\underline{\underline{W}}} \huge \mathcal \orange {\underline{\underline {E}}} \huge \mathcal \purple {\underline{\underline{R}}} ✧$

Given :-

In an equilateral triangle ABC, D is a point on side BC such that BD =1/3BC

To Find :-

AD²

Construction:-

Construct AE perpendicular to BC.

Proof :-

In ️ AEB & ️ AEC

AE = AE [COMMON]

/_AEB = /_AEC [each 90°]

AB = AC [equal sides]

=> ️ AEB  ≅ ️ AEC [RHS Congruency]

Therefore, BE = EC

Now, Let AB = BC = CA = x

So BE = EC = x/2

Now, BD = 1/3 BC = x/3

DE = BE - BD = x/2 - x/3 = x/6.

Now, In triangle ️ AEB

AB² = BE² + AE²

AB² = (BD + DE)² + AE²

AB² = BD² + DE² + 2×BD×DE + AE²

AB² = BD² + 2×BD×DE + (AE² + DE²)

AB² = BD² + 2×BD×DE + AD²

${x}^{2} = {( \frac{x}{3}) }^{2} + 2 \times \frac{x}{3} \times \frac{x}{6} + {AD}^{2} \\ AD² = {x}^{2} - \frac{ {x}^{2} }{9} - \frac{ {x}^{2} }{9} \\ AD² = \frac{9 {x}^{2} - {x}^{2} - {x}^{2} }{9} \\ AD² = \frac{ {7x}^{2} }{9} \\ = > AD² = \frac{7}{9} AB²$

$\red{\textsf{(✪MARK BRAINLIEST✿)}} \\ \huge\fcolorbox{aqua}{aqua}{\red{FolloW ME}}$

Answer 2

Answer:

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Step-by-step explanation:

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