Math, asked by nirxli, 8 months ago

In triangle abc, D is a point on BC such that 2BD=3CD. Prove that 5AB^2 = 5AC^2 + BC^2

Answers

Answered by Anonymous
5

Answer:

hi mate,

Here is your answer

Please mark me as brainleast

Step-by-step explanation:

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.

Answered by Anonymous
0

Answer:

It's right mate which is written above

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