show that $OA\perp DE$ let $\Delta ABC$,and $AB=AD,AC=AE,\angle DAB=\angle CAE$,and $R=DC\bigcap BE$,and the ponit $O$ is $\Delta BRC$ external。show that
$$OA\perp DE$$
 A: Well, there a couple of alternative ways to go about this... Here is one
Let $\angle \, DAB = \angle \, CAE = \pi - 2\theta$. Perform a $\pi - 2\theta$ counterclockwise rotation around point $A$. Then point $D$ is mapped to point $B$ and $C$ is mapped to $E$. Therefore triangle $DAC$ is mapped to triangle $BAE$, so the two triangles are congruent and edge $DC$ is mapped to $BE$, which means that $DC = BE$ and $\angle \, DRB = \pi - 2 \theta = \angle \,DAB$ (where $R = DC \cap BE$). By the way, this means that quad $DBRA$ is inscribed in a circle denoted $k'$. All these argument imply that angle $\angle \, BRC = 2 \theta$.  
Denote by $k$ the circle with center $O$ and passign through the points $B, \, R, \, C$. Let $F$ be the point on the circle $k$ such that $OF$ is orthogonal to chord $BC$ and $F$ is in the same half-plane as $R$ with respect to line $BC$ (see picture). Then $OF$ is the orthogonal bisector of $BC$. 

Angle $\angle \, BRC = \pi - (\pi - 2\theta) = 2 \theta$ and therefore $\angle \, BOC =  2\pi - 4\theta$. Since $OF$ is the orthogonal bisector of $BC$, the two triangles $FOB$ and $COF$ are congruent isosceles triangles with $OB = OF = OC$ and $$\angle \, FOB = \angle \, COF = \frac{1}{2} \, \angle \, BOC = \pi - 2\theta.$$
Consequently the four triangles $DAB, \, CAE, \, FOB$ and $COF$ are all isosceles with angle $\pi-2\theta$ between the equal edges, so they are all similar to each other. 
Since triangles $DAB$ and $FOB$ are similar $$\frac{BD}{BA}=\frac{BF}{BO} = \lambda$$  and one can easily check that $\angle \, DBF = \angle \, ABO$, which means that triangles $DBF$ and $ABO$ are similar, so $$\frac{DF}{AO} = \frac{BD}{BA}=\frac{BF}{BO} = \lambda$$  i.e. $DF = \lambda \, AO$. Analogously, triagnles $ECF$ and $ACO$ are similar and $$\frac{EF}{AO} = \frac{CE}{CA}=\frac{CF}{CO} = \frac{BF}{BO} =  \lambda$$ i.e. $EF = \lambda \, AO$. Thus $DF = EF$ whcih means that triangle $DFE$ is isosceles and if $M$ is the midpoint of $DE$ then $FM$ is ortthogonal to $DE$ and $$\angle \, DFM = \angle \, EFM = \frac{1}{2} \, DFE$$.  
Furthermore, it follows from the similarity of triangles $DBF$ and $ABO$ that $$\angle \, BDF = \angle \, BAO$$ If $K$ is the intersection point of lines $DF$ and $AO$, then $$\angle \, BDK = \angle \, BDF = \angle \, BAO = \angle BAK$$ which in its own turn implies that points $B, D, A, K$ lie on the same circle, which was already denoted by $k'$ (by the way, this means that points $A, D, B, R, K$ all lie on the circle $k'$ but that's not important for the proof). Therefore $$\angle \, DKA = \angle \, DBA = \theta$$
Mreover, again from the similarity of triangles $DBF$ and $ABO$ we see that $$\angle \, DFB = \angle \, AOB$$ Analogously, the similarity of $ECF$ and $ECO$ yields that $$\angle \, EFC = \angle \, AOC$$ Combining the two facts together, we canclude that
$$\angle \, DFB + \angle \, EFC = \angle \, AOB + \angle \, AOC = \angle \, BOC = 2\pi - 4 \theta$$ Now, we can find the angle $$\angle \, DFE = 2\pi - \Big(\angle \, BFC + \angle \, DFB + \angle \, EFC \Big) = 2 \pi - \Big(\angle \, BFC +  2\pi - 4 \theta \Big) = $$
$$= 2 \pi - \Big(2\theta +  2\pi - 4 \theta \Big) = 2\theta$$ which means that from before 
$$\angle \, DFM = \frac{1}{2} \, \angle \, DFE = \theta = \angle \, DKA$$ But the equation
$$\angle \, DFM = \angle \, DKA$$ implies that the lines $AK \equiv AO$ and $MF$ are parallel. However, we have already shown that $MF$ is orthogonal to $DE$ (as a median, angle bisector and altitude in the isosceles triangle $DFE$). 
Thus, $AO$ is orthogonal to $DE$.
