Question

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/2BC. Prove that 9AD² = 7AB².

Answer 1

In △ABE, ∠E = 90° 

⇒ \(\overline{AB}\) is hypotenuse. 

∴ AB² = AE² + BE² 

AE² = AB² – BE² 

⇒ AE² = AB² – (\(\frac{BC}{2}\))²

= AE² = AB² – ( \(\frac{AB}{2}\) )² (∵ AB = BC) 

⇒ AE² = \(\frac{3}{4}\)AB² ……… (1) 

In △ADE, ∠E = 90° 

⇒ \(\overline{AD}\) is hypotenuse. 

⇒ AD² = AE² + DE² 

⇒ AE² = AD² + DE²

⇒ 28 AB² = 36 AD²

⇒ 7 AB² = 9 AD² 

Hence proved.