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