Question

In an equilateral ABC, the side BC is trisected at D. Prove that 9AD2
= 7AB2

Answer 1

Given: In an equilateral triangle ΔABC. The side BC is trisected at D such that BD = (1/3) BC.

To prove: 9AD2 = 7AB2

Construction: Draw AE ⊥ BC.

Proof : In a ΔABC and ΔACE

AB = AC ( Given)

AE = AE ( common)

∠AEB = ∠AEC = 90°

∴ ΔABC ≅ ΔACE ( For RHS criterion)

BE = EC (By C.P.C.T)

BE = EC = BC / 2

In a right angled triangle

ADE AD2 = AE2 + DE2 ---------(1)

In a right angled triangle

ABE AB2 = AE2 + BE2 ---------(2)

From equ (1) and (2)

we obtain ⇒ 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 ΔABC, AB = BC = CA)

⇒ AD2 = AB2 + AB2 / 36 – AB2 / 4

⇒ AD2 = (36AB2 + AB2– 9AB2) / 36

⇒ AD2 = (28AB2) / 36

⇒ AD2 = (7AB2) / 9 9AD2 = 7AB2 .

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Answer 2

Answer:

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