Question

15. In an equilateral triangle ABC, D is a point on side BC such that BD =1/3BC.PTOVE THAT 9AD.AD=7AB.AB​

Answer 1

Given :

• An equilateral triangle ABC
• D is a point on BC
• BD = BC / 3

To prove :

• 9 AD . AD = 7 AB . AB

Construction :

• Draw AM ⊥ BC

Proof :

Refer to the attachment for figure

∵ AM ⊥ BC and ABC is an equilateral triangle

∴ BM = BC / 2   .... eqn (1)

→ DM = BM - BD

→ DM = (BC / 2)  - (BC / 3)

→ DM = (3 BC - 2 BC) / 6

→ DM = BC / 6   ..... eqn (2)

Now,

Consider Δ ABM

by Pythagoras theorem

→ AB² = AM² + BM²

putting AM² = AD² - DM²

→ AB² = AD² - DM² + BM²

using eqn (1) and (2)

→ AB² = AD² - (BC/6)² + (BC/2)²

since , BC & AB are sides of equilateral triangle

→ AB² = AD² - (AB/6)² + (AB/2)²

→ AB² = AD² - AB²/36 + AB²/4

→ AD² = AB² + AB²/36 - AB²/4

→ AD² = (36 AB² + AB² - 9 AB²) / 36

→ AD² = 28 AB² / 36

→ AD² = 7 AB² / 9

cross multiplying

→ 9 AD² = 7 AB²

→ 9 AD . AD = 7 AB . AB

Proved .