In triangle ABC, assume AB<BC and let the angle bisector of ABC intersect AC at point D. Prove that AD < CD

Answer 1

A on the left, C on the right, B on top

let angle lambda = ABD = CDB with h = BD

and x=angle BDA and y = BDC

by law of sines:

sin A/h = sin lambda/ AD = sin x/ AB

and

sin C/h = sin lambda/CD = sin y/BC

the AD = sin lambda* h / sin A

and

CD = sin lambda *h/ sin C

since AB < BC, x < y must hold

So y-x>0

triangles have 180 degrees, so

A+x = C+y

then A = C+y-x > C+ 0 = C

Because A>C and sine in increasing function on [0, 90]

the proof is complete because of the equations above in bold.