Let ABC be an acute angled triangle Let ABC be an acute angled triangle; AD be the bisector of  ∠BAC with D on BC; BE be the altitude from B on AC. Show that ∠CED > 45°
 A: Let us consider the triangle $ABC$. Since the triangle is acute, the altitude from $B$ necessarily intersects $AC$ internally, i.e. $E$ is between $A$ and $C$.

Let the lengths and angles of the triangle be labelled according to the figure above.
From the angular bisector theorem, we have 
$$\frac{c}{b} = \frac{m}{n}$$
where $m = |BD|$ and $n = |CD|$. Likewise, from the generalized angular bisector theorem, we have
$$\frac{h\sin\beta}{k\sin\alpha} = \frac{m}{n}$$
where $h$ is the length of the altitude from $B$ and $k=|CE|$. Note that the claim $\alpha > 45^\circ$ is equivalent to $\alpha > \beta$ since $\alpha + \beta = 90^\circ$. Therefore the claim is true if and only if $\alpha > \beta$ if and only if $\sin\alpha > \sin\beta$. Therefore the problem is equivalent to proving
$$\frac{c}{b} = \frac{h\sin\beta}{k\sin\alpha} < \frac{h}{k}$$
Let $hb = \Delta$ denote the area of the triangle. Then equivalently
$$k < \frac{\Delta}{c} = h_c$$
where $h_c$ is the length of the altitude from $c$. The above inequality is easily seen to be true. Consider swinging the length $CE$ around $C$ to form a circle of radius $k$ centered at $C$. Then the line $BE$ is tangent to the circle at $E$ and clearly separates the circle and the side $AC$. The claim follows.
