15. In an equilateral triangle ABC, D is a point on side BC such that

1.	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» 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 .

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»

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