Solution 1:

Let $\angle BCA = \gamma$ and let $M$ be the point of intersection of lines $OJ$ and $AB$ (Figures 10 and 11).
As $OA = OB$ and $OM$ bisects the angle $AOB$, we have $OM \perp AB$. We also have $\angle AOM = \angle AOB = \angle ACB = \gamma$, because $O$ lies inside the triangle $ABC$. As $\angle AMO = 90^\circ = \angle ADC$, triangles $AMO$ and $ADC$ are similar. Hence also $\angle MAO = \angle DAC = 90^\circ - \gamma$, implying
$$\angle MAJ = \angle JAO = \angle DAE = \angle EAC = \frac{1}{2} (90^\circ - \gamma). \quad (6)$$
Hence triangles $AJO$ and $AEC$ are similar, too. This implies the similarity of triangles $AJE$ and $AOC$ (spiral similarity). As $OA = OC$, the latter similarity implies $JA = JE$. By symmetry in the isosceles triangle $AOB$, we obtain $JA = JB$. Consequently, $JB = JE$. Hence the triangle $JBE$ is isosceles.

Fig. 10

Fig. 11

Solution 2:

Let $\angle CAB = \alpha$, $\angle ABC = \beta$ and $\angle BCA = \gamma$. We start by proving the equalities (6) as in Solution 1. Next, let lines $OJ$ and $BC$ meet in $F$, whereas let lines $AJ$ and $BC$ meet in $X$ (Figures 12 and 13 depict situations that differ by the order of points $E$ and $F$). We show that points $A, J, E, F$ are concyclic. From equalities (6) we get
$$\begin{aligned} \angle JAE &= \angle BAC - \angle BAJ - \angle EAC \\ &= \alpha - \frac{1}{2}(90^\circ - \gamma) - \frac{1}{2}(90^\circ - \gamma) \\ &= \alpha - (90^\circ - \gamma) = \alpha + \gamma - 90^\circ = 180^\circ - \beta - 90^\circ = 90^\circ - \beta. \end{aligned}$$
As $OA = OB$, the line $OJ$ is the perpendicular bisector of the line segment $AB$. Thus
$$\angle XFJ = \angle BFJ = 90^\circ - \angle ABF = 90^\circ - \beta.$$

Fig. 12

Fig. 13
Altogether, we have $\angle XAE = \angle JAE = 90^\circ - \beta = \angle XFJ$, implying that $A, J, E, F$ are concyclic.

Therefore $\angle JEA = \angle JFA = \angle JFB = \angle JFX = \angle XAE = \angle JAE$, implying $JA = JE$. But $JA = JB$ as $J$ lies on the perpendicular bisector of the line segment $AB$. Consequently, $JB = JE$. Hence the triangle $JBE$ is isosceles.

Solution 3:

We use the known fact that lines drawn from a vertex of a triangle to its orthocentre and its circumcentre are isogonals which implies that the angles between these lines and the respective sides are equal.
As $AD$ and $AO$ are the altitudes and the circumradius of the triangle $ABC$, both drawn from vertex $A$, we have $\angle BAO = \angle CAD$. But $AE$ bisects the angle $CAD$ and $J$ is the point of intersection of angle bisectors of the triangle $ABO$, hence $\angle BAJ = \angle CAD = \angle EAD$. This shows that rays $AJ$ and $AD$ are isogonals in the triangle $ABE$. As $AD$ is also an altitude of the triangle $ABE$, the line $AJ$ must pass through the circumcentre of the triangle $ABE$. But the circumcentre of the triangle $ABE$ must also lie on the perpendicular bisector of the side $AB$. As $OA = OB$, the bisector of the angle $AOB$ coincides with the perpendicular bisector of the side $AB$. Hence the circumcentre of the triangle $ABE$ is the point $J$ of intersection of lines $AJ$ and $OJ$. Consequently, $JB = JE$, implying that the triangle $JBE$ is isosceles.

Solution 4:

As $OA = OB$, the bisector of the angle $AOB$ coincides with the perpendicular bisector of the side $AB$. Hence the point $J$ lies on the perpendicular bisector of the side $AB$, implying that points $A$ and $B$ lie on some circle $\omega$ with centre $J$.
We show that $\angle AJB = 2\angle AEB$. Indeed,
$$\begin{aligned} \angle AEB &= \angle AED = 90^\circ - \angle EAD \\ &= 90^\circ - \frac{\angle CAD}{2} = 90^\circ - \frac{90^\circ - \angle ACD}{2} = 90^\circ + \frac{\angle ACD}{2}. \end{aligned}$$
As $JA = JB$ implies $\angle JAB = \angle JBA$, we obtain
$$\begin{aligned} \angle AJB &= 180^\circ - 2\angle JAB = 180^\circ - \angle OAB \\ &= 180^\circ - \frac{180^\circ - \angle AOB}{2} = 90^\circ + \frac{\angle AOB}{2} = 90^\circ + \angle ACB = 2\angle AEB. \end{aligned}$$

Hence $\omega$ is the circumcircle of the triangle $ABE$. Consequently, $JB = JE$, implying that the triangle $JBE$ is isosceles.