In ∆ABC, seg AD ⊥ seg BC, DB = 3CD. Prove that: 2AB2 = 2AC2 + BC2

Explanation:- 

In right angled triangle ADB, applying Pythagoras theorem, we get:

AB^2 = AD^2 + DB^2

In right angled triangle ADC, applying Pythagoras theorem, we get:

AC^2 = AD^2 + CD^2

Adding the above two equations, we get:

AB^2 + AC^2 = 2AD^2 + DB^2 + CD^2

We need to express DB and CD in terms of BC.

From the given condition, we have:

DB = 3CD

Therefore, CD = DB/3

Also, since AD is perpendicular to BC, we have:

AD^2 = BD * DC

Substituting the values of CD and DB, we get:

AD^2 = (3CD) * CD = 3CD^2

Substituting these values in the above equation, we get:

AB^2 + AC^2 = 2(3CD^2) + DB^2 + CD^2

Simplifying this equation using the given condition, we get:

AB^2 + AC^2 = 8/3(DB^2 + CD^2)

But we also know that:

DB^2 + CD^2 = BC^2

Substituting this value, we get:

AB^2 + AC^2 = 8/3(BC^2)

Simplifying, we get:

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

Hence, we have proved that 2AB^2 = 2AC^2 + BC^2, given that in triangle ABC, AD is perpendicular to BC and DB is three times as long as CD.