Question
In \triangle ABC, segment CD bisects \angle C and intersects AB at point D. Show that CD^2 < CA\cdotp CB
Answer 1
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
Answer 2
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
Answer 3
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