Nsryjdtyk

I am trying to use sparql to find all French labels of words that are related to a given English word.

The query I made is something like the following:

SELECT ?extword ?word ?label where {

    ?word rdfs:label "shark"@eng.

    {?extword rdf:type ?word}

    UNION

    {?word rdf:type ?extword}.

    ?extword rdfs:label ?label filter(lang(?label)="fra")

    }

While it does give me the French labels of the extword as desired, it does not give the French labels of word itself.

What I need is to insert a clause to express the relation that extword is exactly word, like this:

SELECT ?extword ?word ?label where {

    ?word rdfs:label "shark"@eng.

    {?extword = ?word}

    UNION

    {?extword rdf:type ?word}

    UNION

    {?word rdf:type ?extword}.

    ?extword rdfs:label ?label filter(lang(?label)="fra")

    }

It turns out to be syntax error. Is it possible to also include the French labels of word itself, without doing another query?

To be clearer, the result I expect is the result from my first query combined with result from the following query:

SELECT ?word ?label where {

    ?word rdfs:label "shark"@eng.

    ?word rdfs:label ?label filter(lang(?label)="fra")

    }