#### Question

In \triangle ABC, segment CD bisects \angle C and intersects AB at point D. Show that CD^2 < CA\cdotp CB

Plot an auxiliary line from point D which intersects AC at point E to make \angle CDE = \angle DBC

Since CD is the bisector of \angle C

\angle BCD = \angle ACD

\triangle CBD \sim \triangle CDE

\dfrac{CD}{BC}=\dfrac{CE}{CD}

CD^2=CB\cdotp CE < CB \cdotp CA

Plot circumcircle of \triangle ABC . Extend CD to make it intersect the circle at point E.

Since \angle CBD and \angle AEC are subtended by the same arc AC,

\angle CBD = \angle AEC

Since CD is the bisector of \angle C

\angle BCD = \angle ACD

\triangle CBD \sim \triangle CAD

\dfrac{CB}{CD} =\dfrac{CE}{CA}

CA\cdotp CB = CD\cdotp CE =CD\cdotp (CD+ED) =CD^2+CD\cdotp ED

According to Intersecting Chords Theorem,

CD\cdotp ED = AD\cdotp BD

Therefore,

CD^2 = CA\cdotp CB-AD\cdotp BD < CA\cdotp CB

Let a=CB, b=CA, m= CD, \alpha = \dfrac{ \angle C}{2}

Using the Law of Cosines,

BD^2= a^2+m^2-2am\cos \alpha

(1)

AD^2 = b^2+m^2-2bm\cos \alpha

(2)

According to Angle bisector theorem,

\dfrac{CA}{CB} = \dfrac{AD}{BD}

Square the equation

\dfrac{CA^2}{CB^2} = \dfrac{AD^2}{BD^2} =\dfrac{b^2}{a^2}

Plug in (1) and (2)

\dfrac{b^2+m^2-2bm\cos \alpha}{a^2+m^2-2am\cos \alpha} =\dfrac{b^2}{a^2}

a^2b^2+m^2a^2-2a^2bm\cos \alpha=a^2b^2+m^2b^2-2ab^2m\cos \alpha

m(a^2-b^2)=2ab(a-b)\cos \alpha

m=\dfrac{2ab\cos \alpha}{a+b}

\because \cos\alpha <1

\therefore m<\dfrac{2ab}{a+b}

Square the inequality

m^2 < \dfrac{4a^2b^2}{(a+b)^2} =ab\cdotp \dfrac{4ab}{(a+b)^2}< ab

Now we have verified the inequality

CD^2 < CA\cdotp CB