Question

In ∆ABC, AD ⊥ BC and point D lies on BC such that 2DB = 3CD. Prove that 5AB2 = 5AC2+ BC2

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Answer 1

Since ⊿ADB is a right triangle, we have  

AB² = AD² + DB².  

And since 2DB =3 CD, we know BC = BD + CD BC = 2/3 DB+DB  And DB = (3/5) CB.  

(1) AB² = AD² + 9/25 BC2.  

Similarly ⊿ADC is a right triangle, so  

AC² = AD² + DC²,  

So Similarly, DC = BC - BD DC= BC (2/5), and  

(2) AC² = AD² + (4/25) BC²,  

Subtract (1) by (2)  

AB² - AC² = (9 - 4)/25 BC² AB² - AC² = 1/5 BC² 5AB² - 5AC² = BC²

So  

5 AB² = 5 AC² + BC².

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Answer 2

Answer:

Since ⊿ADB is a right triangle, we have  

AB² = AD² + DB².  

And since 2DB =3 CD, we know BC = BD + CD BC = 2/3 DB+DB  And DB = (3/5) CB.  

(1) AB² = AD² + 9/25 BC2.  

Similarly ⊿ADC is a right triangle, so  

AC² = AD² + DC²,  

So Similarly, DC = BC - BD DC= BC (2/5), and  

(2) AC² = AD² + (4/25) BC²,  

Subtract (1) by (2)  

AB² - AC² = (9 - 4)/25 BC² AB² - AC² = 1/5 BC² 5AB² - 5AC² = BC²

So  

5 AB² = 5 AC² + BC².

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Step-by-step explanation: