In $\triangle ABC$, $AD$ is the altitude and $CE$ is the bisector. Find $\angle EDB$ if $\angle CEA = 45^\circ$.

Question

Recently I took part in a mathematics Olympiad. There were 25 questions for 2 hours, and I couldn’t solve this problem, I’ll be glad if you help me with this:

In $\triangle ABC$, $AD$ is the altitude and $CE$ is the bisector. Find $\angle EDB$ if $\angle CEA = 45^\circ$.

this is 9 class olympic, task ball is 1.5

Answer 1

In the figure, $F$ is a point on $BC$ such that $CF=CA$.

Join the lines as shown.

Note that $\Delta CEF \cong \Delta CEA $ (SAS).

$\therefore \angle CEF=\angle CEA=45^{\text o}$ and hence $\angle AEF=90^{\text o}.$

Together with $AE=EF, $ we have $\angle EAF=45^{\text o}.$

Note also that $\angle ADC= \angle AEF=90^{\text o}$.

$\therefore A, D, F, E$ are concyclic.

Hence $\angle EDF=\angle EAF=45^{\text o}.$

$\therefore \angle EDB=45^{\text o}.$

Answer 2

As can be seen in figure:

$\angle CEA=45^o=\frac 12(\overset {\large\frown} {AL}+\overset {\large\frown} {BM})$

the extension of DE meets the circle at G and we have:

$LM||GA\Rightarrow \overset {\large\frown} {AL}=\overset {\large\frown} {MG}$

which deduces $\angle EDB=\angle CEA=45^o$