On an acute triangle $ABC$ we draw the internal bisector of $<ABC$, $BE$, and the altitude $AD$, ($D$ on $BC$), show that $<CDE$ it's bigger than 45 degrees.
Problem
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Tags: trigonometry, geometry, incenter, circumcircle, angle bisector, geometry unsolved
professordad
02.08.2013 00:54
Note that $\frac{\sin{\angle CAD}}{\cos{\angle CAD}} < \frac{\sin{\angle BAC}}{\cos{\angle CAD}}$. But $\cos{\angle CAD} = \sin{\angle ACD}$, and $\frac{\sin{\angle BAC}}{\sin{\angle ACD}} = \frac{BC}{AB}$ by LOS. Also, $\frac{\sin{\angle CAD}}{\cos{\angle CAD}} = \frac{CD}{AD}$. So
\[\frac{CD}{AD} < \frac{BC}{AB}\]
By Angle Bisector theorem, $\frac{BC}{AB} = \frac{CE}{AE}$. So
\[\frac{CD}{AD} < \frac{CE}{AE}\]
Now imagine that there is a point $F$ on $AC$ such that $DF$ is the angle bisector of $\angle ADC$. By Angle Bisector Theorem, $\frac{CD}{AD} = \frac{CF}{AF}$. So then
\[\frac{CF}{AF} < \frac{CE}{AE}\] It follows that $F$ must be between $C$ and $E$. Since $\angle CDF = 45^{\circ}$, $\angle CDE > 45^{\circ}$.
MexicOMM
02.08.2013 01:19
professordad wrote:
Note that $\frac{\sin{\angle CAD}}{\cos{\angle CAD}} < \frac{\sin{\angle BAC}}{\cos{\angle CAD}}$. But $\cos{\angle CAD} = \sin{\angle ACD}$, and $\frac{\sin{\angle BAC}}{\sin{\angle ACD}} = \frac{BC}{AB}$ by LOS. Also, $\frac{\sin{\angle CAD}}{\cos{\angle CAD}} = \frac{CD}{AD}$. So
\[\frac{CD}{AD} < \frac{BC}{AB}\]
By Angle Bisector theorem, $\frac{BC}{AB} = \frac{CE}{AE}$. So
\[\frac{CD}{AD} < \frac{CE}{AE}\]
Now imagine that there is a point $F$ on $AC$ such that $DF$ is the angle bisector of $\angle ADC$. By Angle Bisector Theorem, $\frac{CD}{AD} = \frac{CF}{AF}$. So then
\[\frac{CF}{AF} < \frac{CE}{AE}\] It follows that $F$ must be between $C$ and $E$. Since $\angle CDF = 45^{\circ}$, $\angle CDE > 45^{\circ}$.
I was hoping on a simpler solution since this is a problem 4.
Arab
02.08.2013 10:19
Since $\triangle ABC$ is acute,we get $\frac{1}{2}\angle ABC+\angle ACB>\frac{1}{2}(\angle ABC+\angle ACB)>45^\circ$,so $\angle AEI>\angle ADI$,where $I$ is the incenter of $\triangle ABD$,which implies that $E$ lies within the circumcircle of $\triangle ADI$,which follows $\angle ADE=90^\circ-\angle CDE<\angle AIE=45^\circ$,hence $\angle CDE>45^\circ$. $Q.E.D.$
Tsikaloudakis
02.08.2013 20:59
$\begin{array}{l} {\rm{Distinguish }}\mathop {}\limits^{} {\rm{the }}\mathop {}\limits^{} {\rm{cases:}}\\ \\ \alpha .{\rm{ }}\mathop {}\limits^{} \mathop {}\nolimits^{} {\rm{if}}\mathop {}\limits^{} {\rm{ \hat {\rm B}}} \ge {\rm{4}}{{\rm{5}}^o}\mathop {}\limits^{} {\rm{ }}{\rm{, then: }}\hat \varphi > {\rm{\hat {\rm B}}}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \hat \varphi > {45^o}\mathop {}\limits^{} {\rm{ }}({\rm{ }}\mathop {}\limits^{} {\rm{as}}\mathop {}\limits^{} {\rm{ external}}\mathop {}\limits^{} {\rm{ angle}}\mathop {}\limits^{} {\rm{ triangle}}\mathop {}\limits^{} \mathop {}\limits^{} {\rm B}D{\rm I})\\ \\ {\rm{b}}{\rm{.}}\\ {\rm{ }}\mathop {}\limits^{} {\rm{ }}\left. \begin{array}{l} {\rm{ \hat {\rm B}}} < {\rm{4}}{{\rm{5}}^o}\\ \\ {\rm{AB}} \le {\rm{BC}}\\ \\ {\rm{AM = MC}} \end{array} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \left. \begin{array}{l} \hat \theta < {45^o}\\ \\ {\rm E}C > EA \end{array} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \hat \varphi > \hat \omega > {45^o}\\ \\ c.\\ \left. \begin{array}{l} {\rm{ }}\mathop {}\limits^{} {\rm{\hat {\rm B}}} < {\rm{4}}{{\rm{5}}^o}\\ {\rm{ }}\mathop {}\limits^{} {\rm{AB}} > {\rm{BC}}\\ \\ {\rm{ }}\mathop {}\limits^{} {\rm{BA = BZ}} \end{array} \right\}\mathop {}\limits^{} \mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \hat \varphi > \hat \theta > {45^o}\\ {\rm{ }} \end{array}$
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Math_CYCR
21.10.2015 04:55
Generalization.- Let $ABC$ be an acute-angled triangle.$D$ lies on $BC$ with $D$ between $B, C$. Let $BE$ be the angle bisector of $\angle ABC$ with $E$ on $AC$. Prove that $\angle CDE > \frac{\angle CDA}{2}$ Proof.- Suppose this isn`t the case so there is a point $F$ on the segment $AE$ ($F \neq A$) such that $DF$ is the angle bisector of $\angle CDA$. Notice that in $\triangle ABC \longrightarrow \frac{\sin \angle BAC}{\sin \angle ACB} = \frac{BC}{AB} = \frac{CE}{AE}$ Notice that in $\triangle ADC \longrightarrow \frac{\sin \angle DAC}{\sin \angle ACD} = \frac{DC}{AF} = \frac{CF}{AF}$ $\longrightarrow \frac{CF}{AF} \ge \frac{CE}{AE} \longrightarrow \frac{\sin \angle DAC}{\sin \angle ACD} \ge \frac{\sin \angle BAC}{\sin \angle ACB} = \frac{BC}{AB} \longrightarrow \sin \angle DAC \ge \sin \angle BAC$ which is a contradiction since $\angle BAC < 90$. So we can conclude, that $F$ lies on the segment $EC$ with $F$ between $E, C$ and obtain $\angle CDE > \frac{\angle ADC}{2}$
neelanjan
21.10.2015 15:11
draw a perpendicular on AB .and try by it congruences and similarity . basically , this is a problem of RMO- 2007 . please follow the link :http://olympiads.hbcse.tifr.res.in/uploads/rmo-2007 to get a better solution .