Question

The perpendicupar from A on side BC of a triangle ABC intersects bc at d such that db=3cd prove that 2qb square= 2ac square + bc square.

Answer 1

Let BD = 3y , CD = y, and thus BC = 4y.
Applying Pythagoras theorem,

In ∆ABD , AB² = AD²+BD²

In ∆ACD, AC² = AD²+CD²,

from equations above equate for AD, we get,

AB²–BD² = AC²–CD²

AB² = AC² + 9y²–y²
= AC² + 8y²
multiply by 2,

2AB² = 2AC²+16y²
which is,

2AB² = 2AC² + BC²