Question

In an equilateral triangle ABC, D is the point on side BC such that BD=1/3BC. Prove that 9AD^2=7AB^2​

Answer 1

$\huge{\mathfrak{Question:-}}$

In an equilateral triangle ABC, D is the point on side BC such that BD = 1/3BC. Prove that 9AD² = 7AB²

$\huge{\mathfrak{Answer:-}}$

Given:- ABC is a equilateral triangle. D is the point on the side such as BD = 1/3BC.

Proof:

In Δ ABE and Δ ACE

AB = AC    (Given)

AE = AE    (Common)

∠AEB = ∠AEC  (90°)

∴ ΔABE ≅ ΔACE     (By RHS Congruence criterion)

⇒ BE = EC    (c.p.c.t)

$\bold{= BE = EC = \frac{BC}{2}}\\ \\ \\ \bold{In\;right \triangle ADE,}\\ \\ \bold{AD^{2}=AE^{2}+DE^{2}}\;\;\;\;\;\;\;\;............(1) \;\;\;\;\; \bold{(Pythagorus\;theorem)}\\ \\ \bold{In\; right \triangle ABE,}\\ \\ \bold{AB^{2}=AE^{2}+BE^{2}}\;\;\;\;\;\;\;\;............(2) \;\;\;\;\ \bold{(Pythagorus\;theorem)}\\ \\ \bold{From\;equation\;(1)\;and(2),\;we\;get}\\ \\ \bold{AD^{2}-AB^{2}=DE^{2}-BE^{2}}\\ \\ \bold{\implies AD^{2}-AB^{2} = (BE-BD)^{2}-BE^{2}}$

$\bold{\implies AD^{2}-AB^{2} = (\frac{BC}{2}-\frac{BC}{3})^{2}-(\frac{BC}{2})^{2}\;\;\;\;\;\;\;\; (\therefore BD = \frac{1}{3}BC,\;BE = \frac{1}{2}BC)}\\ \\ \\ \bold{\implies AD^{2}-AB^{2} = (\frac{3BC-2BC}{6})^{2}-(\frac{BC}{2})^{2}}\\ \\ \\ \bold{\implies AD^{2}=AB^{2}+\frac{BC^{2}}{36}-\frac{BC^{2}}{4}}\\ \\ \\ \bold{\implies AD^{2}=AB^{2}+\frac{AB^{2}}{36}-\frac{AB^{2}}{4} \;\;\;\;\;\;\;\;\; (AB=BC=CA)}\\ \\ \\ \bold{\implies AD^{2}=\frac{36AB^{2}+AB^{2}-9AB^{2}}{36}}$

$\bold{\implies AD^{2}=\frac{28AB^{2}}{36}}\\ \\ \\ \bold{\implies AD^{2}=\frac{7}{9}AB^{2}}$

⇒ 9 AD² = 7 AB²

Answer 2

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step-by-step explanation:

Given,

A ∆ABC in which AB = BC = CA

and

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

Required To prove :

9AD² = 7AB² .

Construction :

Draw AL ⊥ BC .

Proof :

In right triangles ALB and ALC, we have

AB = AC ( given )

AL = AL ( common )

Angle ALB = Angle ALC = 90°

∴ ∆ALB ≅ ∆ ALC

[ By RHS congruence criteria ] .

So,

BL = CL ( by CPCT)

Thus,

BD = ⅓BC

and

BL = ½BC .

Now,

In ∆ALB, ∠ALB = 90° .

∴ AB² = AL² + BL² .......(1)

[ by Pythagoras' theorem ] .

In ∆ALD , ∠ALD = 90° .

∴ AD² = AL² + DL²

[ 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✍️✍️