Question

In an equilateral triangle ABC the side BC us trisected at D. Prove that 9AD^2=7AB^2

Answer 1

Answer:

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 ;- In a ΔABC and ΔACE

AB = AC ( given)

AE = AE (common)

∠AEB = ∠AEC = (Right angle)

∴ ΔABC ≅ ΔACE By RHS Creition

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 ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎

‎‎Its proved!!!