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Question:
In an EQUILATERAL triangle ABC, D is a point on side BC such that BD =1/3 BC. PROVED that 9AD^2=7AB^2.
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◀ 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
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
Its proved!!!
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◀ 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
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
Its proved!!!
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bhoomibasrani:
didi
Answered by
63
hey mate
here's the proof
here's the proof
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