In an equilateral triangle ABC , D is A point on side BC such that BD = 1/3BC.Prove that 9( AD) ^ 2= 7( AB) ^ 2
Answers
Given that :-
In Δ ABC is an equilateral triangle.
D is point on BC such that BD =BC.
To prove:-
9 AD² = 7 AB²
Construction: Draw AE ⊥ BC.
Proof :-
In a ΔABC and ΔACE
AB = AC ( given)
AE = AE (common)
∠AEB = ∠AEC = (Right angle)
Hence ΔABC ≅ ΔACE (By RHS Creation)
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 )
( we know that 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
= ) 9AD² = 7AB² proved
Hope it helps you
Answer:
Step-by-step explanation:
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 Creition
∴ ΔABC ≅ ΔACE
Again,
BE = EC (By C.P.C.T)
BE = EC = BC 2
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
= ) 9AD2 = 7AB2
Hence, 9AD2 = 7AB2
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