Question

ABC is an equilateral triangle, P is a point such that BP : PC = 2:1 . Prove that :
9 AP² = 7AB²

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Answer 1

Answer:

It is given that in an equilateral triangle ΔABC, The side BC is trisected at P,

such that BP = (1/3) BC

=> Draw AE ⊥ BC.

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 APE

AP² = AE² + PE² -(1)

In a right angled triangle ABE

AB² = AE² + BE² -(2)

From equation (1) and (2) we get,

=> AP² - AB² = PE³ - BE²

=> AP² - AB² = (BE – BP)² - BE²

=> AP² - AB² = (BC/2 – BC/3)² – (BC/2)²

=> AP² - AB² = ((3BC – 2BC)/6)² – (BC/2)²

=>AP² - AB² = BC²/36 – BC²/4 ( In a equilateral triangle ΔABC, AB = BC = CA)

=> AP²= AB² + AB²/36 – AB²/4

=> AP² = (36AB² + AB²– 9AB²)/36

=> AP² = (28AB²)/36

=> AP² = (7AB²)/9

[9AP² = 7AB²]

Answer 2

Answer:

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