Question

In an equilateral triangle ABC, D is a point on side BC such that BD=1/3BC. prove that 9ADsquare=7ABsquare​

Answer 1

Correct question:-

• In an equilateral triangle ABC, D is a point on side BC such that BD=⅓BC. prove that 9AD²=7AB².

Given:-

• Equilateral triangle ABC D is a point an BC.

To prove:-

• 9AD²=7AB².

Construct:-

• Lets draw AE _|_ BC.

Proof:-

All sides of equilateral triangle is equal.

$\sf \: AB=BC=AC$

$\sf \: Let \: AB=BC=AC = x$

Given

$\sf \: BD = \frac{1}{3}BC$

$\sf \: BD = \frac{x}{3}$

In ∆AEB and ∆AEC

• AE=AE (common)
• AB=AC. ( Both x as it is equilateral triangle)
• ∠AEB= ∠AEC. ( both 90° as AE_|_ BC)

Hence by RHS congruency

$∆AEB \cong ∆AEC$

$\sf \: \therefore \: BE= EC \: \: \: \: \: \: \: \: \: \: \: ( By \: CPCT )$

• So, BE=EC= 1/2 BC
• BE=EC=x/2

So,

$\sf \: BE = \frac{ x }{2}$

$\sf \: BD + DE = \frac{x}{2}$

$\sf \: \frac{x}{3} + DE = \frac{x}{2}$

$\sf \: DE = \frac{x}{2} - \frac{x}{3}$

$\sf \: DE = \frac{3x - 2x}{6}$

$\sf \: DE = \frac{x}{6}$

Using pythagoras theorem

⇒(Hypotenuse)²= (Height)²+ (Base)²

Now in right ∆ AEB

⇒AB²=AE²+(BE)²

$\sf \: ⇒x²= (AE)² +(\frac{x}{2})^2$

$\sf \: ⇒x² =(AE)² + (\frac{x}{2})^2$

$\sf \: ⇒x²-\frac{x}{4}^2=AE²$

$\sf \: AE ^{2} = \frac{3x^{2} }{4} \: \: \: \: \: \: \: \: \: (1)$

Similarly,

In right ∆AED

$\sf \: AD²=AE²+DE²$

$\sf \: AD² = {3x}^{2} + ( \frac{x }{6} )^{2}$

$\sf \: AD²= \frac{3 {x}^{2} }{4} + \frac{ {x}^{2} }{36}$

$\sf \: AD²= \frac{( {3x}^{2}) \times 9 \times {x}^{2} }{36}$

$\sf \: AD²= \frac{27 {x}^{2} + {x}^{2} }{36}$

$\sf \: AD²= \frac{28 {x}^{2} }{36}$

$\sf \: AD²= \frac{7 {x}^{2} }{9}$

$\sf \: 9AD²=7 {x}^{2}$

$\sf \: 9AD²=7AB ^{2}$

Hence proved.

Answer 2

⠀⠀⠀⠀⠀⠀⠀${\huge{\underbrace{\rm{Correct~Question}}}}$

• In an equilateral triangle ABC, D is a point on side BC such that BD=⅓BC. prove that 9AD²=7AB².

⠀⠀⠀⠀⠀⠀⠀⠀${\huge{\underbrace{\rm{Required~Answer}}}}$

Given:-

• A ∆ABC in which AB = BC = CA

• D is a point on BC such that BD = ⅓BC.

To prove :-

• 9AD² = 7AB² .

Construction :-

• Draw AL ⊥ BC .

Proof :-

In right triangles ALB and ALC, we have

AB = AC $\bf\pink{(Given)}$

AL = AL $\bf\red{(Common)}$

Angle ALB = Angle ALC = 90°

∴ ∆ALB ≅ ∆ ALC

$\bf\blue{ [ By ~RHS ~congruence~ criteria ~] .}$

So,

BL = CL ( by CPCT)

Thus,

BD = ⅓BC

and

BL = ½BC .

Now,

In ∆ALB, ∠ALB = 90° .

∴ AB² = AL² + BL²..(1)

$\bf\blue{ [ by~ Pythagoras'~ theorem~ ] .}$

In ∆ALD , ∠ALD = 90° .

∴ AD² = AL² + DL²

$\bf\blue{ [ by~ Pythagoras'~ theorem~ ] .}$

⇒ AD² = AL² + ( BL - BD )² .

⇒ AD² = AL² + BL² + BD² - 2BL.BD .

⇒ AD² = ( AL² + BL² ) + BD² - 2BL.BD .

⇒ AD² = AB² + BD² - 2BL.BD. [ using (1) ]

⇒ AD² = BC² + ( ⅓BC )² - 2( ½BC ). ⅓BC .

[ ∵ AB = BC, BD = ⅓BC and BL = ½BC ] .

⇒ AD² = BC² + 1/9BC² - ⅓BC² .

⇒ AD² = 7/9BC² .

⇒ AD² = 7/9AB² [ ∵ BC = AB ] .

=> 9AD² = 7AB²

★ Hence, proved ★