in an equilateral ABC,D is a point on side BC such that BD=1/2BC prove that 9AD^2=7AB^2
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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 .
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 .
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sankalppatra7:
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Your answer is in photo
I have done this early
So hope it helps you
I have done this early
So hope it helps you
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