Question

in triangle ABC, AD perpendicular to BC, DB=3CD.prove that, 2AB^2=2AC^2+BC^2

Answer 1

BC= BD+DC= 3CD+DC= 4CD
BC²=(4CD)²=16CD²
In triangle ADC,
AC²=AD²+DC²
AD²= AC²-DC²
In triangle ABD,
AB²=AD²+BD²
AB²=AC²-DC²+(3CD)²
AB²=AC²-DC²+9DC²
AB²=AC²+8CD²
Multiplying both the sides by 2,
2AB²=2AC²+16CD²
2AB²=2AC²+BC²

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 - 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.

Step-by-step explanation: