Question

❤️BRAINLY CHALLENGE ❤️

In triangle ABC, AD is perpendicular to BC and point D lies on BC such that 2DB=3CD. Prove that:

5AB^2=5AC^2+BC^2

❤️50 point ❤️

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Answer 1

Answer :-

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Answer:

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 - BDDC= 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.

Answer 2

$\sf\huge{\underline{\mathfrak{A_nswer}}}$

▶$\sf\huge{\underline{Question!}}$

➡In triangle ABC, AD is perpendicular to BC and point D lies on BC such that 2DB=3CD. Prove that:

5AB^2=5AC^2+BC^2

$\sf\bold{\underline{P_rooF!}}$

Given: Consider ∆ABD,

such that,

$AB^{2} = AD^{2} + BD^{2}$==(1)

we have!

2DB = 3CD

Then, $CD = \frac{2\times DB}{3}$

from Figure!

BC = BD + CD

BC = BD + $\frac{2\times BD}{3}$

BC = $\frac{5\times BD}{3}$

BD = $\frac{3\times BC}{5}$

Substituting in (1)

$AB^{2} = AD^{2} + BD^{2}$

$AB^{2} = AD^{2} +\frac{3^{2}\times BC^{2}}{5^{2}}$

$AB^{2} = AD^{2} +\frac{9\times BC^{2}}{25}$==(2)

Similarly, in ∆ACD

we have

$AC^{2} = AD^{2} + DC^{2}$==(3)

From fig!

CD = BC - BD

since, BD = $\frac{3}{2}CD$

then..

CD = $BC -\frac{3}{2}CD$

BC = $CD + \frac{3}{2}CD$

BC = $\frac{5}{2}CD$

DC = $\frac{2}{5}BC$

substituting in (3)

$AC^{2} = AD^{2} + DC^{2}$

$AC^{2} = AD^{2}+\frac{2^{2}}{5^{2}}BC^{2}$

$AC^{2} = AD^{2}+\frac{4}{25}BC^{2}$==(4)

subtracting (4) from (2)

$AC^{2} = AD^{2}+\frac{4}{25}BC^{2}$

$AB^{2} = AD^{2} +\frac{9\times BC^{2}}{25}$

$AC^{2}-AB^{2}=\frac{4-9}{25}BC^{2}$

$AC^{2} - AB^{2} =\frac{-5}{25}BC^{2}$

$AC^{2} - AB^{2} =\frac{-1}{5}BC^{2}$

$AB^{2} - AC^{2} =\frac{1}{5}BC^{2}$

$5(AB^{2} - AC^{2})= BC^{2}$

$5\times AB^{2}-5\times AC^{2}= BC^{2}$

$5\times AB^{2} =5\times AC^{2} + BC^{2}$

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