Question

In triangle ABC, AD prependicular to BC such that 2DB=3CD prove 5AB^=5AC^+ BBC

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 ) mark as brilliant

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: