in equilateral triangle ABC,D is a point on side BC such that BD =1÷3BC.Prove that 9AD.AD=7AB.AB.
Answers
BE= EC=BC/2= a/2..
and AE = a√3/2
BD= 1/3 BC ( GIVEN)
BD= a/3
DE=BE - BD =a/2 - a/3 = a/6
applying Pythagoras theorem in triangle ADE we get...
AD.AD = (a√3/2×a√3/2) + ( a/6× a/6)....
AD.AD = (3a× a/4) + (a×a / 36)
AD×AD= ( 2a×a/36)
AD×AD= 7/9 AB × AB
9AD×AD= 7AB× AB .....
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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