# ABC is an acute angled triangle. The bisector of /_ BAC intersects BC at D. BE is a perpendicular drawn on AC from B. Points E and D are joined. Show that /_CED>45^o How to show?

Apr 30, 2016

Some thoughts to begin an answer...

#### Explanation:

OK, here's a diagram to start:

Let:

$\left\{\begin{matrix}A = \left(0 0\right) \\ B = \left(e h\right) \\ C = \left(c 0\right) \\ D = \left(f k\right) \\ E = \left(e 0\right) \\ F = \left(f 0\right)\end{matrix}\right.$

Then:

$\sin \theta = \frac{D F}{A D} = \frac{k}{\sqrt{{f}^{2} + {k}^{2}}}$

$\cos \theta = \frac{A F}{A D} = \frac{f}{\sqrt{{f}^{2} + {k}^{2}}}$

$\sin 2 \theta = \frac{B E}{A B} = \frac{h}{\sqrt{{e}^{2} + {h}^{2}}}$

$\cos 2 \theta = \frac{A E}{A B} = \frac{e}{\sqrt{{e}^{2} + {h}^{2}}}$

From the double angle formulae we have:

$\frac{e}{\sqrt{{e}^{2} + {h}^{2}}} = \cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta = \frac{{f}^{2} - {k}^{2}}{{f}^{2} + {k}^{2}}$

$\frac{h}{\sqrt{{e}^{2} + {h}^{2}}} = \sin 2 \theta = 2 \cos \theta \sin \theta = \frac{2 f k}{{f}^{2} + {k}^{2}}$

Since $D$ lies on $B C$ there is some $t \in \left[0 , 1\right]$ such that:

$\left(f , k\right) = \left(c \left(1 - t\right) + e t , h t\right)$

We want to show that if $A B C$ is acute then $D F > E F$.

In other words, $k > f - e$.

Since the triangle is acute, there are some other conditions we can throw in:

$e > 0$

$e < c$

$A {B}^{2} + B {C}^{2} > A {C}^{2}$, i.e. $\sqrt{{e}^{2} + {h}^{2}} + \sqrt{{\left(c - e\right)}^{2} + {h}^{2}} > {c}^{2}$

May 1, 2016

In acute angled$\Delta B A C$,$\angle B A C < {90}^{o}$ and $\angle C A D \mathmr{and} \angle E A D = \frac{1}{2} \angle B A C$ ,AD being bisector of angle $\angle B A C$
So$\angle C A D < {45}^{O}$
In$\Delta E A O ,$ $\angle A E O = {90}^{o} , \angle E A O < {45}^{o}$,hence$\angle A O E > {45}^{o}$
$\therefore A E > O E$
EP is taken equal in length of OE then P,O are joined and produced.
E,D are also joined and produced ,which meets the produced PO at Q.
Now in $\Delta P O E , P E = O E \mathmr{and} \angle P E O = {90}^{o}$
So $\angle E P O = \angle E O P = {45}^{o}$

Now in $\Delta P Q E$,
the exterior$\angle C E Q = \angle C E D = \angle E P Q + \angle P Q E$
$\implies \angle C E D = {45}^{o} + \angle P Q E$ $\text{Since} \angle E P Q = \angle E P O = {45}^{o}$

Hence $\angle C E D > {45}^{o}$ Proved