Please help me 7th march is my maths exam solve this question
In an equilateral triangle ABC , D is a point on side BC, such that BD =1/3 BC. Prove that 9AD square = 7 AB square.
Answers
Answer:
Given = BD = 1/3 BC
To prove = 9AD^2 = 7AB^2
CONSTRUCTION = We draw AE ⊥ BC
Proof :- Let the sides of the equilateral triangle be a and AE be the altitude of
Δ ABC
∴ BE = EC = BC/2 = a/2
And AE = a √3/2
ALSO,
BD = 1/3BC
∴ DE = BE-BD
⇒ a/2 -a/3 = a/6
BY APPLYING PYTHAGORAS THEOREM
AD^2 = AE^2+DC^2
AD^2 = (a √3/2)^2+ (a/6)
AD^2 = (3a^2/4) + (a/36)
AD^2 = 28a^2/36
AD^2 = 7AB^2/9
9AD^2 = 7AB^2
HOPE IT HELPS YOU
Diagram in attachment .
Given that ABC is an equilateral triangle .
All sides are equal .
Construct AE .
∠AED = ∠AEC = 90° .
In Δ AEB and Δ AEC ,
AB = AC
AE = AE
∠AEB = ∠AEC
Δ AEB ≅ Δ AEC [ RHS ]
⇒ EB = EC [ cpct ]
By Pythagoras theorem :
In Δ AEB ,
AE² = AB² - EB²
⇒ AE² = AB² - ( BC/2 )²
⇒ AE² = AB² - ( AB/2 )² [ ∵ AB = BC ]
⇒ AE² = AB² - AB²/4
⇒ AE² = 3 AB²/4
By Pythagoras theorem :
In Δ AED :
AD² = AE² + ED²
⇒ AD² = 3 AB²/4 + ED²
ED = BC - 1/2 BC - 1/3 BC [ ∵ BE = EC ]
⇒ ED = 1/2 BC - 1/3 BC
⇒ ED = 1/6 BC
⇒ AD² = 3 AB²/4 + ( BC/6 )²
⇒ AD² = 3 AB²/4 + ( AB/6 )² [ ∵ AB = BC ]
⇒ AD² = 3 AB²/4 + AB²/36
⇒ AD² = ( 27 AB² + AB² ) / 36
⇒ 36 AD² = 28 AB²
⇒ 9 AD² =7 AB² .
Hence proved !
