Question

The perpendicular from A on side BC of a triangle ABC intersects BC at D such that BD = 3CD. Prove that 2 AB 2 2 AC 2 = BC 2

Answer 1

We have two right angle triangles: ΔABD and  Δ ACD.  We are given 
   BD = 3 CD    -- (1)
So  BC = BD + CD = 4 CD   -- -(2)

From Pythagoras theorem : 
         AB² = AD² + BD²    and    AC² = AD² + CD²

Adding these two equations, we get :

AB² - AC² = BD² - CD²
                = (3 CD)²  - CD²       using (1)
                = 8 CD²

2 (AB² - AC²) = 16 CD²
                       = (4 CD)² = BC²      using (2)

Answer 2

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