# The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that $$2{AB}^2$$ = $$2{AC}^2$$ + $${BC}^2$$

## Solution :

We have : DB = 3CD

Now, BC = DB + CD

i.e. BC = 3CD + CD          [because BD = 3CD]

BC = 4CD

$$\therefore$$   CD = $$1\over 4$$ BC  and  DB = 3CD = $$3\over 4$$ BC             ……….(1)

Since triangle ABD is a right triangle, right angled at D, therefore, by Pythagoras Theorem, we have :

$${AB}^2$$ = $${AD}^2$$ + $${DB}^2$$             ……….(2)

In triangle ACD,

$$\angle$$ D = 90 ,  $${AC}^2$$ = $${AD}^2$$ + $${CD}^2$$              ……..(3)

Subtracting (3) from (2), we get

$${AB}^2$$ – $${AC}^2$$ = $${DB}^2$$ – $${CD}^2$$

$${AB}^2$$ – $${AC}^2$$ = $$({3\over 4}BC)^2$$ – $$({1\over 4}BC)^2$$              (using (1))

$${AB}^2$$ – $${AC}^2$$ = $$({9\over 16} – {1\over 16})$$$${BC}^2$$

$${AB}^2$$ – $${AC}^2$$ = $${1\over 2}$$$${BC}^2$$

$$\implies$$  $$2{AB}^2$$ = $$2{AC}^2$$ + $${BC}^2$$