$\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$

Question

If bisector of angle C of an acute triangle ABC cuts the side AB in D and the circumcircle of triangle ABC in E,then $\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$.Then what is value of k?

Since angle bisector of C,bisects the arc BC.And circumcenter O will fall on line CE.And $\angle AOB=2\angle C$.These concepts i know but i caanot relate them to get answer.maybe i am lacking somewhere.Any suggestions or guidance is welcome.

Answer 1

$\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$. What is value of $k$?

Note that \begin{align} \angle EAB&=\angle EBA=\tfrac\gamma2 \end{align}

Assuming that $|AB|=c$, $|BC|=a$, $|CA|=b$,
$\angle BAC=\alpha$, $\angle ABA=\beta$, $\angle ACB=\gamma$, we can find $|CD|$ and $|CE|$ from the area relations:

\begin{align} S_{\triangle ABC}&= S_{\triangle ADC}+ S_{\triangle BDC} \\ S_{CAEB}&= S_{\triangle ABC}+S_{\triangle AEB} = S_{\triangle ACE}+S_{\triangle BCE} \\ S_{\triangle AEB}&= \tfrac14 c^2\tan\tfrac\gamma2 \end{align}

which gives

\begin{align} |CD| &= \frac{ab\sin\gamma}{(a+b)\sin\tfrac\gamma2} \\ |CE| &= \frac{ab\sin\gamma+\tfrac12 c^2\tan\tfrac\gamma2}{(a+b)\sin\tfrac\gamma2} \\ |DE|&=|CE|-|CD| = \frac{\tfrac12 c^2\tan\tfrac\gamma2}{(a+b)\sin\tfrac\gamma2} \\ \frac{|CE|}{|DE|} &= \frac{2ab\sin\gamma}{c^2\tan\tfrac\gamma2}+1 = \frac{2ab(\cos\gamma+1)}{c^2}+1 \\ &= \frac{a^2+b^2-c^2+2ab +c^2}{c^2} = \frac{(a+b)^2}{c^2}. \end{align} Hence $k=1$.

Edit:

Note that $|CE|$ is a common side of $\triangle BCE$ and $\triangle ACE$:

\begin{align} S_{BCE}+S_{ACE}&=S_{ABC}+S_{AEB} \\ \tfrac12|CE|a\sin\tfrac\gamma2 + \tfrac12|CE|b\sin\tfrac\gamma2 &= \tfrac12ab\sin\gamma + \tfrac14c^2\tan\tfrac\gamma2 \\ |CE|\sin\tfrac\gamma2(a+b) &= ab\sin\gamma + \tfrac12c^2\tan\tfrac\gamma2 \\ |CE|&= \frac{ab\sin\gamma + \tfrac12c^2\tan\tfrac\gamma2}{\sin\tfrac\gamma2(a+b)}. \end{align}

Edit 2:

In $\triangle AEB$ we have $\angle EAB=\angle EBA=\tfrac\gamma2$, thus it's isosceles and $|AE|=|BE|$. Its area $S_{AEB}=\tfrac12 h |AB|$, where $h$ is the altitude, which splits isosceles $\triangle AEB$ into two congruent right triangles with the base $\tfrac12c$, so $h=\tfrac12|AB|\tan\angle EAB= \tfrac12c \tan\tfrac\gamma2$, hence $S_{AEB}=\tfrac12 h c=\tfrac14c^2 \tan\tfrac\gamma2$.