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STEP $1$:
Join $E$,$D$
.$AO$ & $ED$ intersects at $F$
According to the question
$\angle{BAC}=\angle{ABC}=\angle{ACB}=60$

By easy angle chasing we get,

$\angle{ECB}=20^{\circ}$

$\angle{ABD}=20^{\circ}$

$\angle{DBC}=40^{\circ}$

$\angle{ECA}=40^{\circ}$

$\angle{BOC}=120^{\circ}$

$\angle{DOC}=60^{\circ}$

$\angle{BDC}=80^{\circ}$

$\angle{BEC}=80^{\circ}$

$\angle{ADO}=100^{\circ}$

$\angle{EOD}=120^{\circ}$

$\angle{EOD}+\angle{EAD}=180^{\circ}$

So $OEAD$ is a inscribed in a circle.
So,$\angle{EDO}=\angle{EAO}$

STEP $2$:

Using cosine law,$\angle {ADE}=90^{\circ}$
So,
$\angle{EDO}=10^{\circ}$

$\angle{EAO}=10^{\circ}$

$\angle{OAD}=50^{\circ}$

$\angle{OAD}+\angle{ACO}=90^{\circ}$

$\angle{AOC}=90^{\circ}$