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°
[Math] Let ABC be an acute angled triangle
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Related Solutions
As in the above drawing let $\angle DAL=\angle OAC=a$
And let be the intersection of the angle bisector, perpendicular bisector and the altitude be $O$
Let the perpendicular bisector be $DK$
$\angle DAO=\angle OAH=a and \angle AOH=\angle DOA=90-a$
Then $\triangle DAO \equiv \triangle OAH (A.A.S)$
With this you can get that $AD=AH$ and $OD=OH$
Since $DK$ is the perpendicular bisector of $AB$ $AD=DB$1
Now draw $RH$ and also the median respective to the right angle in a right angles triangle is equal to the half of the hypotenuse
$AD=AH$
And that brings us to $AD=AH=DH$
Since $\triangle ADH$ is equilateral
$\angle BAC=\angle ADH=\angle DHA=60$
Best Answer
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.