Let ABC be a triangle with a, b, and c the lengths of the
sides opposite vertices A, B, and C respectively.
Let the cevian CD bisect ∠ACB.

Prove the following:

    |CD|² = ab[ 1-{c/(a+b)}² ]

without using the Law of Cosines.

Let X be the foot of the perpendicular from C to BA, and let x be the
displacement of X from A, measured in the direction BA. Let |XC| = h.

Since CD bisects angle ACB, |AD| = bc/(a + b).

In triangle AXC:             b² = x² + h²                               (1)

In triangle BXC:             a² = (x + c)² + h²                       (2)

In triangle DXC:             |CD|² = (x + bc/(a + b))² + h²     (3)

(2) – (1) =>      a² – b² = 2cx + c²                                   (4)

(3) – (1) =>      |CD|² – b² = 2bcx/(a + b) + {bc/(a + b)}²

Using (4) now gives:

   |CD|²             = b² + b(a² – b² –c²)/(a + b) + {bc/(a + b)}²

            = [b²(a + b)² + b(a + b)(a² – b² – c²) + b²c²]/(a + b)²

            = ab[a² + b² + 2ab – c²]/(a + b)²

            = ab[1 – {c/(a + b)}²]

Hopefully, no one will notice that (1) & (2) together constitute the Cosine Rule!

Posted by Harry on 2013-10-08 17:00:24