Question

In triangle ABC, AD perpendicular to BC and point D lies on BC such that 2DB=3CD. Prove that 5AB squre=5AC square+BC square.

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 - 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.

:) Hope this helps!!!!!

Answer 2

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.