Ex 6.5, 15

Transcript

Ex 6.5, 15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7 AB2 Given: Equilateral triangle ABC D is a point an BC Such that BD = 1/3 BC To prove: 9 AD2 = 7 AB2 Construction: Lets draw AE ⊥BC Proof: All sides of equilateral triangle is equal, AB = BC = AC Let AB = BC = AC = x Given BD = 1/3BC BD = 𝑥/3 In ∆ 𝐴𝐸𝐵 𝑎𝑛𝑑 ∆ 𝐴𝐸𝐶 AE = AE AB = AC ∠ 𝐴𝐸𝐵=∠𝐴𝐸𝐶 Hence by RHS congruency ∆ 𝐴𝐸𝐵 ≅ ∆ 𝐴𝐸𝐶 ∴ BE = EC So, BE = EC = 1/2BC BE = EC = 𝑥/2 So, BE = 𝑥/2 BD + DE = 𝑥/2 𝑥/3+𝐷𝐸=𝑥/2 DE = 𝑥/2−𝑥/3 DE = (3𝑥 − 2𝑥)/(2 × 3) DE = 𝑥/6 Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 Similarly In right ∆ 𝐴𝐸𝐷 AD2 = AE2 + DE2 AD2 = 3𝑥2/4+( 𝑥/6 )^2 AD2 = 3𝑥2/4+𝑥2/36 AD2 = ((3𝑥2) × 9 + 𝑥2)/36 AD2 = (27𝑥2 + 𝑥2)/36 AD2 = (28 𝑥2)/36 AD2 = 7𝑥2/9 9AD2 = 7x2 9 AD2 = 7 AB2 Hence proved