Math, asked by angel152, 1 year ago

in an equilateral triangle ABC,D is a point on the side BC such that BD=1/3BC .prove that 9AD^=7AB^.

Answers

Answered by Anonymous
15

Given: ΔABC is an equilateral triangle. D is point on BC such that BD =BC.


To prove: 9 AD² = 7 AB²


Construction: Draw AE ⊥ BC.


Proof ;-


Considering on Triangles which are given below;-


In a ΔABC and ΔACE


AB = AC ( given)


AE = AE (common)


∠AEB = ∠AEC = (Right angle)


∴ ΔABC ≅ ΔACE


By RHS Creation


∴ ΔABC ≅ ΔACE


Considering On Question;-


Again,


BE = EC (By C.P.C.T)


BE = EC = BC²


In a right angled ΔADE


AD²= AE2 + DE² ---(1)


In a right angled ΔABE


AB² = AE² + BE² ---(2)


From equation (1) and (2) ;


 =) AD²  - AB² =  DE² - BE².


 =) AD²  - AB² = (BE – BD)² - BE².


 = ) AD²  - AB² = (BC / 2 – BC/3)² – (BC/2)²


 = AD2  - AB2 = ((3BC – 2BC/6)² – (BC/2)² 


 = AD²  - AB² = (BC² / 36 – BC2 / 4 )


( In a equilateral triangle, All sides are equal to each other)


AB = BC = AC


 = ) AD²= AB² + AB²/ 36 – AB² / 4


 = )AD² = (36AB² + AB²– 9AB²) / 36


 = ) AD² = (28AB²) / 36


=) AD² = (7AB²) / 9


Cross Multiplication here,


= ) 9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎

‎Hence, 9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎

‎Its proved!!!



Answered by Anonymous
3

Step-by-step explanation:

➡ Given :-

→ A ∆ABC in which AB = BC = CA and 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 ( given ) and AL = AL ( common )

∴ ∆ALB ≅ ∆ ALC [ By RHS axiom ] .

So, BL = CL .

Thus, BD = ⅓BC and BL = ½BC .

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 ] . ....

 \huge \green{ \boxed{ \sf \therefore 9AD^{2} = 7AB^{2}. }}

Hence, it is proved.

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