Question

In an equilateral triangle ABC,D is point on BC such that BD=1/3BC.Prove that 9AD²=7AB².

Answer 1

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Let the side of an equilateral triangle be a, and AE be the altitude of ΔABC.
∴BE=EC=BC2=a2
And, AE=a√32
given that, BD=13BC
∴BD=a3
DE=BE−BD=a2−a3=a6
Applying Pythagoras theorem in ΔADE, we get
AD2=AE2+DE2
⇒ AD2= (a√32)2+(a6)2
⇒ AD2=(3a24)+(a236)
⇒AD2=28a236
⇒ AD2=79AB2 [Since, the side of an equilateral triangle is a.i.e;AB=BC=AC=a]
⇒9AD2=7AB2