Question

in ΔABC AD_I_BC and point D lies on BC such that 2DB = 3CD prove that 5AB^2=5AC^2+BC^2

Answer 1

We have a ∆ ABC , where AD perpendicular to BC  , And  2 DB  =  3 CD




SO,

DBCD = 32

SO,

DB  = 33 + 2BC   = 35BC

And

CD  = 23 + 2BC   = 25BC

Now we apply Pythagoras theorem in ∆ ABD ,  we get

AB2  =  AD2  + DB2

AB2  =  AD2  + (35BC) 2                                           ( As given DB  = 35BC )

AB2  =  AD2  + 925BC 2
   
Taking LCM , we get

25 AB2  =  25 AD2  + 9 BC2  ------------------------- ( 1 )

And

Now we apply Pythagoras theorem in ∆ ACD ,  we get

AC2  =  AD2  + CD2

AC2  =  AD2  + (25BC) 2                                           ( As given CD  = 25BC )

AC2  =  AD2  + 425BC 2
   
Taking LCM , we get

25 AC2  =  25 AD2  + 4 BC2  ------------------------- ( 2 )

Now we subtract equation 2 from equation 1 and get

⇒25 AB2  - 25 AC2  =  25 AD2  + 9 BC2  - 25 AD2  - 4 BC2   

⇒25 AB2  - 25 AC2  = 5 BC2 

⇒5 ( 5 AB2  - 5 AC2  ) = 5 BC2 

⇒5 AB2  - 5 AC2  =   BC2 

⇒5 AB2  = 5 AC2  +  BC2  ( Hence proved )

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.