Bed angle

Let $F$ be the foot of the perpendicular from $B$ to $AD$ . Then $BD=CD=\sqrt7$ and $FD=DE=\sqrt3$ . Since $D$ is the midpoint of $BC$ , $D$ is the midpoint of $EF$ . Then $EF = 2DE=2\sqrt3$ .

By using Pythagoras' theorem, we obtain $BF = \sqrt{BD^2-DF^2} = 2$ . Therefore $\tan \widehat{BED} = \frac{BF}{EF} = \frac{1}{\sqrt3}$ . Then $\widehat{BED} = 30^\circ$ .

No matter how you draw the diagram (regardless of whether E and A are on the same side of BDC), the following solution will work.

In right angled CED, $CD = \sqrt{7}$ and $ED = \sqrt{3}$ , so Pythagoras yields $CE = 2$ . That gives the sine and cosine of CDE to be $\frac{2}{\sqrt{7}}$ and $\frac{\sqrt{3}}{\sqrt{7}}$ respectively.

So the sine and cosine of BDE are $\frac{2}{\sqrt{7}}$ and $-\frac{\sqrt{3}}{\sqrt{7}}$ respectively.

Then, apply the cosine rule in triangle BDE, around angle D, to obtain BE = 4. Then, apply the sine rule, using sides BD and BE, to get sin(BED) = $\frac{1}{2}$ .

Argue that since BEC is not reflex, it is less than 180 to obtain BED less than 90. So BED = 30.

$ED$ is a median of $\bigtriangleup EBC$ so

$DE^2=\frac{EB^2+EC^2}{2}-\frac{BC^2}{4}$

$\Leftrightarrow DE^2=\frac{EB^2+(DE^2+DC^2)}{2}-\frac{(2DC)^2}{4}$

(because $\bigtriangleup EDC$ is a right triangle at $D$ )

$\Leftrightarrow EB=4.$

$\bigtriangleup EDC$ has $\cos \widehat{BED}=\frac{ED^2+EB^2-BD^2}{2\cdot EB\cdot ED}=\frac{16+3-7}{2\cdot 4\cdot \sqrt{3}}=\frac{\sqrt{3}}{2}$

Hence, $\widehat{BED}=30^{\circ}.$

Consider the triangle $\triangle{BEC}$ with cevian $DE$ . Also note that by Pythagorean Theorem (since $\triangle{CED}$ is right) that $CE^2 + DE^2 = CD^2 \implies CE^2 + 3 = 7 \implies CE=2$ . Additionally, $BD=CD=\sqrt{7}$ .

Then by Stewart's Theorem, we have that $CE^2 \cdot BD + BE^2 \cdot CD = DE^2 \cdot BC + BD \cdot CD \cdot BC$ Using the ascertained values above, we find that $4\sqrt{7} + BE^2 \sqrt{7} = 6\sqrt{7} + 14 \sqrt{7} \implies BE^2 = 16 \implies BE=4$

Now consider triangle $\triangle{BED}$ . Let $\angle{BED} = \theta$ . Then since $BE=4, BD=\sqrt{7}, ED = \sqrt{3}$ , by Law of Cosines we have $BD^2 = BE^2+DE^2 - 2 BE \cdot DE \cdot \cos\theta \implies 7=16+3-8\cos\theta\sqrt{3}$ and solving for $\cos \theta$ yields $\cos \theta = \dfrac{\sqrt{3}}{2} \implies \theta = \boxed{30^{\circ}}$ .

Clearly $CE = \sqrt {DC^{2}-DE^{2}} = \sqrt {7-3} = 2$
Let $BE = a$ . Also, $BC= 2DC= 2\sqrt 7$ , $DE= \sqrt 3$ .

Thus by Appolonius Theorem in triangle $BEC$ with median $DE$ , we get $a^{2}+2^{2} = 2[(\sqrt 7)^{2} + (\sqrt 3)^2]$ , And thus we get $a=4$ . So $BE=4$ .

Clearly $BE^{2}+CE^{2}- 2(BE)(CE)(cos 120^ \circ)$ = $4^{2} + 2^{2} - 2(4)(2)(-0.5)$ = $28 = BC^{2}= BE^{2} + CE^{2} -2(BE)(CE)(cos \angle BEC)$ , due to Cosine Rule.

Therefore $\angle BEC = 120^\circ = \angle BED + \angle CED$ , but $\angle CED=90^\circ$ .

Therefore $\angle BED = 30^\circ$ .

Since $DC =\sqrt{7}$ then $BD = \sqrt{7}$ ,

and according to Pythagorean Theorem, $CE^2 + ED^2 = DC^2$ which gives the value $EC=2$

so $cos \angle EDC = \sqrt\frac{3}{7}$ since $\angle EDB = 180^\circ -\angle EDB$ , then $cos \angle EDB = -\sqrt\frac{3}{7}$

By using Cosine Rule, $BD^2 + DE^2 - 2 \times BD \times DE \times cos \angle EDB = BE^2$

resulting in $BE=4$

Then use the Sine Rule, $\frac{BE}{sin(\angle EDB)} = \frac{BD}{sin(\angle BED)}$ giving the value of $sin \angle BED = \frac{1}{2}$ which implies that $\angle BED= 30^\circ$ since it is an acute angle

Draw the diagram. Actually, the case for $E$ inside and outside of $\triangle ABC$ are same. And we see that,

$BD = DC = \sqrt{7}$

And, from $\triangle CDE$ , we obtain that : $\cos \angle EDC = \frac{DE}{DC} = \frac{\sqrt{3}}{\sqrt{7}} \implies \cos \angle BDE = \cos(180 - \angle EDC) = - \cos \angle EDC = \frac{-\sqrt{3}}{\sqrt{7}}$

Now, see the triangle $BDE$ . Now, by cosinus law, we obtain that : $BE^2 = BD^2 + DE^2 - 2. BD. DE. \cos \angle BDE = 7 + 3 - 2. \sqrt{21}. \frac{-\sqrt{3}}{\sqrt{7}} = 16 \implies BE = 4$

also,

$\cos \angle BED = \frac{BE^2 + DE^2 - BD^2}{2. BE. ED} = \frac{16 + 3 - 7}{2. 4. \sqrt{3}} = \frac{\sqrt{3}}{2} \implies \angle BED = \boxed{30^{\circ}}$

Notice that: CD=DB=sqrt(7); <ADB=180-ADC. We can note in the triangle CDE that cos(<EDC)=sqrt(3/7). Applying the law of cosins in the triangle BDE we can conclude that EB=4 because we know the length of DE and DB and that cos(<EDB)=cos(180 - <EDC)= -sqrt(3/7). Applying again the law of cosins in the triangle BDE we can conclude that cos(<BED)=sqrt(3)/2 (we know the lengths of all the three sides of the triangle, so we can know the angle between any of them) with that we conclude that <BED= 30.

$CE=2, BF=2, BE=4, \angle BED=30$

Let $F$ be the point on $AD$ such that $BF \perp AD$ . Since $D$ is the midpoint, hence $BFD$ and $CED$ are congruent triangles. It follows that $BF = CE = \sqrt{ 7 - 3} = 2$ and $EF = 2 ED = 2 \sqrt{3}$ . Hence, $\tan \angle BEF = \frac{ BF} { FE} = \frac{ 1} { \sqrt{3}}$ . Thus, $\angle BED = \angle BEF = 30 ^\circ$ .

Prolong AD over a distance DE and name this point E'. Now EE' and BC are the diagonals of a parallellogram intersecting in D. Triangle EE'B has a perpendicular angle in E'. Perpendicular sides are 2*root(3) and 2. Tan(BED) = 1/root(3) and thus BED is 30°.