Question

Let $ABC$ be an acute-angled triangle; $AD$ be the bisector of $\angle$ $BAC$ with $D$ on $BC$; and $BE$ be the altitude from $B$ on $AC$. Find $\angle$ $CED$

• A. $>45$ ${}^{\circ }$
• B. $<45$ ${}^{\circ }$
• C. $=30$ ${}^{\circ }$
• D. None

Answer 1

The correct option is A $>45$ ${}^{\circ }$

Draw DL perpendicular to AB; DK perpendicular to AC; and DM perpendicular to BE. Then EM $=$ DK.
Since AD bisects $\angle$A, we observe that $\angle$ BAD $=\angle$KAD. Thus in triangles ALD and AKD, we see that $\angle$LAD $=\angle$KAD.
$\angle$AKD $=$ 90${}^{\circ }=\angle$ALD; and AD is common. Hence triangles ALD and AKD are congruent, giving DL $=$ DK. But DL > DM, since BE lies inside the triangle (by acuteness property). Thus EM > DM.
This implies that $\angle EDM>\angle DEM={90}^{\circ }-\angle EDM.$
We conclude that $\angle EDM>{45}^{\circ }$
Since $\angle CED=\angle EDM$, the results follows.