In an equilateral Δ ABC, D is a point on the side BC, such that BD = 1/2BC. Prove that 9AD² = 7AB²
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Given: In an equilateral triangle ΔABC. The side BC is trisected at D such that BD = (1/3) BC.
To prove: 9AD2 = 7AB2
Construction: Draw AE ⊥ BC.
Proof :
In a ΔABC and ΔACE
AB = AC ( Given)
AE = AE ( common)
∠AEB = ∠AEC = 90°
∴ ΔABC ≅ ΔACE ( For RHS criterion)
BE = EC (By C.P.C.T)
BE = EC = BC / 2
In a right angled triangle ADE
AD2 = AE2 + DE2 ---------(1)
In a right angled triangle ABE
AB2 = AE2 + BE2 ---------(2)
From equ (1) and (2) we obtain
⇒ 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 ΔABC, AB = BC = CA)
⇒ AD2 = AB2 + AB2 / 36 – AB2 / 4
⇒ AD2 = (36AB2 + AB2– 9AB2) / 36
⇒ AD2 = (28AB2) / 36
⇒ AD2 = (7AB2) / 9
9AD2 = 7AB2 .
Hope it helps U
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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